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Calculation of derivatives will learn the power rule, the product rule, the reciprocal
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rule, and the quotient rule for finding derivatives.
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We begin by noting the derivatives of the few basic functions.
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First, any constant function f of x equals c has derivative f prime of x equals zero.
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First think of the graph of constant function that is a horizontal line, which always has
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slope zero.
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The identity function f of x equals x has a derivative f prime of x equals one.
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So, again, just think of the graph of y equals x, which is a straight line, which always
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has slope one.
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The power rule.
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If n is a positive integer, then the function f of x equals x to the nth power has derivative
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f prime of x equals n times x to the n minus one.
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In other words, to find the derivative of f prime, simply multiply by the power and reduce
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the power by one.
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For example, the derivative of x squared is two x to the first power.
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The derivative of x to the fifth is five x to the fourth power.
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Now, let's try to see why the power rule works.
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The difference quotient for x to the nth power is x plus h to the n minus x to the n divided
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by h.
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We can expand the first term here using the binomial theorem, obtaining x to the nth power
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plus remaining terms, each of which has at least one factor of h.
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Because we have x to the n minus x to the n, so all the remaining terms in the numerator
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involve at least one factor of h.
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So we can divide h into each of those terms, and the first term we have is n times x to
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the n minus one, and all of the other terms still involve at least one factor of h.
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Therefore, as h goes to zero, we are left with only the first term n times x to the
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n minus one.
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Linearity properties.
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The derivative of the function f plus g is f prime plus g prime.
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In other words, the derivative of sum is the sum of the derivatives.
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To see why this is true, let's look at the difference quotient for f plus g, f of x plus
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h plus g of x plus h minus f of x plus g of x all divided by h.
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Now we simply rearrange terms here in the numerator, and then split up into two fractions.
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We simply have the sum of the difference quotients for f and g.
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Therefore as h goes to zero, we obtain f prime of x plus g prime of x.
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Two, the derivative of a constant times f is the constant times the derivative of f.
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To see why this is true, let's look at the difference quotient for c times f.
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c times f of x plus h minus c times f of x divided by h.
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And notice that we can factor out a c, and we simply have c times the difference quotient
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for f, which approaches c times f prime of x as h goes to zero.
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For example, if f of x is x squared plus x cubed, then f prime of x can be computed by
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simply taking the derivative term by term.
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The derivative of x squared is 2x, and the derivative of x cubed is 3x squared using the
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power rule, so the derivative of x squared plus x cubed is 2x plus 3x squared.
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The derivative of 3x to the fourth is simply 3 times the derivative of x to the fourth,
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which is 3 times 4x cubed or 12x cubed.
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Here's an example that combines both rules.
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The derivative of 3x squared minus x is 3 times the derivative of x squared minus the
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derivative of x or 6x minus 1.
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Example, let's find p prime of x if p of x is the polynomial x to the fifth plus 2x
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cubed minus 5x squared plus 3x plus 2.
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We'll find p prime by simply taking the derivative p term by term.
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The derivative of x to the fifth is 5x to the fourth.
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Then we add 2 times the derivative of x cubed, which is 3x squared, and then minus 5 times
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the derivative of x squared, which is 2x, and plus 3 times the derivative of x, which is
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1, and then plus the derivative of 2, which is 0.
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So this simplifies, as you see here, 5x to the fourth plus 6x squared minus 10x plus
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3.
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Another example.
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The p of x equal 2x cubed minus 3x squared minus 12x plus 5.
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We want to find all x where p prime of x is positive, and all x where p prime of x is
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negative.
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We'll first calculate the derivative.
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It's 2 times the derivative of x cubed, which is 3x squared minus 3 times the derivative
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of x squared, which is 2x minus 12 times the derivative of x, which is 1, plus the derivative
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of 5, which is 0.
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So we have 6x squared minus 6x minus 12.
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Now since we want to know where this is positive and negative, we'll factor, first factoring
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out the common factor 6, and then factoring x squared minus x minus 2 into x plus 1 times
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x minus 2.
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So we can see that the derivative of p prime of x is 0 at x equals minus 1 and x equals
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2.
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Moreover, p prime of x is negative when x is between minus 1 and 2.
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Think of the graph of the quadratic function here.
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It's a parabola that opens upward, so it must be negative between its 2 zeros.
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And p prime of x is positive, but x less than minus 1, as well as 4x greater than 2.
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Now let's think about what this means in terms of the graph of the function p of x.
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Notice that the graph of p of x has positive slope when x is less than minus 1, negative
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slope when x is between minus 1 and 2, and positive slope when x is greater than 2.
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And this is exactly what we learned from determining where the derivative is positive and where
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the derivative is negative.
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Here's another derivative of a basic function.
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The square root function f of x equals square root of x has derivative f prime of x equals
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1 over 2 times the square root of x.
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This is a straightforward calculation.
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Let's look at the difference quotient, the square root of x plus h minus the square root
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of x divided by h.
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We multiply the top and bottom of this fraction by the square root of x plus h plus the square
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root of x, multiplying out the two numerators, we obtain x plus h minus x, or simply h,
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and then we can cancel an h to obtain 1 over the square root of x plus h plus the square
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root of x.
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And clearly, as h goes to 0, this quotient approaches 1 over 2 times the square root
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of x for all positive x, of course.
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Example, let's find and simplify up prime of x if f of x is 3 times the square root of
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x minus x squared.
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We'll first write the derivative of the first term, which will be 3 times the derivative
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of the square root of x, or 3 times 1 over 2 times the square root of x, and then the
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second term gives us minus 2x using the power rule.
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Now simplification here will essentially be a matter of combining the two terms together
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over a common denominator.
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To do that, we multiply the second term by 2 times the square root of x over 2 times
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the square root of x, and now combining the terms gives us 3 minus 4x times the square
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root of x divided by 2 times the square root of x.
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The product rule.
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The derivative of the product f times g is the derivative of f times g plus f times
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the derivative of g.
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For example, the derivative of x times the square root of x is the derivative of x, which
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is 1 times the square root of x, plus x times the derivative of the square root of x.
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we obtain the square root of x plus 1 half the square root of x or 3 halfs the square root
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of x. Now let's see if we can see why the product rule is true. The difference quotient
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or f of x times g of x is f of x plus h times g of x plus h minus f of x times g of x
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all divided by h. Now we use a little bit of trickery here. We'll make some space between
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the terms in the numerator and subtract f of x times g of x plus h and add the same term.
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Now we group the first two terms and the last two terms together and factor g of x plus
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h from the first two terms in the numerator and factor f of x from the last two terms
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in the numerator. Now we break the fraction apart into two terms. The first term is the
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difference quotient for f times g of x plus h. The second term is f of x times the difference
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quotient for g. Now as h goes to zero, first term approaches f prime of x times g of x
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and the second term approaches f of x times g prime of x. Example, let f of x equal
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x squared minus 1 times the square root of x. Find and simplify f prime of x. So we
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apply the product rule here. f prime of x will be the derivative of x squared minus 1
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or 2x minus 0 times the square root of x plus the first factor x squared minus 1 times
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the derivative of the second factor 1 over 2 times the square root of x. So the first
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term is just 2x times the square root of x. The second term is x squared minus 1 divided
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by 2 square root of x. Now we multiply the first term by 2 times the square root of x over
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2 times the square root of x to obtain a common denominator. And now the first term is 4x
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squared over 2 times the square root of x. And now adding, we obtain 5x squared minus
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1 divided by 2 times the square root of x. Example, let f of x equal the product of 2
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polynomials x squared minus x times x cubed plus x squared minus x plus 1. And let's find
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f prime of 1. Now we'll use the product rule to calculate the derivative. f prime of x
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equals the derivative of the first factor 2x minus 1 times the second factor plus the
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first factor times the derivative of the second factor, which is 3x squared plus 2x minus
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1 plus 0. Now since we only want to know f prime of 1, we may as well simply plug in x
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equals 1 and just do the calculation. The final result here is 2. Example, let f of x equal
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the square of x cubed plus x minus 3, find and simplify f prime of x. To use the product
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rule here, we'll write f of x as x cubed plus x minus 3 times itself. Now f prime is
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the derivative of the first factor 3x squared plus 1 times the second factor plus the first
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factor times the derivative of the second factor, which is again 3x squared plus 1. So we
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have two identical terms here. And so f prime of x is 2 times x cubed plus x minus 3 times
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3x squared plus 1. Example, let f of x equal the square of the function g of x. Let's
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find f prime of x in terms of g of x and g prime of x. And notice that the previous
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example is just a special case of this. Just as before, we'll write f of x as g of x times
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g of x and use the product rule, the derivative of the first factor, g prime of x times the
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second g of x plus the first factor times the derivative of the second factor. Again,
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both terms are the same and so we have two times g of x times g prime of x, which is exactly
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the form of the result in the previous example. The reciprocal rule, the derivative of 1 over
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g is minus g prime divided by g squared. Let's look at a quick derivation of this result.
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Let f of x equal 1 over g of x and we'd like to find f prime of x. We'll multiply both
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sides by g of x and then take the derivative of both sides of that equation. The derivative
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of the left hand side by the product rule is f prime of x times g of x plus f of x times
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g prime of x. And the derivative of the constant 1 is of course 0. Now, we'll solve for f prime
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of x by first subtracting f of x times g prime of x from both sides and rewrite f of x as
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1 over g of x. Now, we'll divide by g of x to obtain f prime of x equals minus g prime
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of x divided by g of x squared. Now, for example, let's let f of x equal 1 over x squared
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plus 1. Let's find f prime of x. The reciprocal rule tells us that the denominator will be
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the square of the denominator of f and the numerator will be minus the derivative of
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the denominator. So we have minus 2x divided by x squared plus 1 squared. Second example.
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Let f of x equal 1 over the square root of x. Find and simplify f prime of x. Again,
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using the reciprocal rule, we have minus a fraction whose denominator is the square
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of the square root of x. Whose numerator is the derivative of the denominator of f.
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And the derivative of the square root of x recall is 1 over 2 times the square root of
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x. Now, we simplify and get minus 1 over 2x times the square root of x. Now, this could
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be written in different ways, none of which are particularly simpler than the others.
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We could write minus 1 over 2 times x to the 3 half's power. Or we could rationalize
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the denominator and write minus the square root of x over 2x squared.
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Reciprocal powers. If anyone is a positive integer, then f of x equals 1 over x to the
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m as the derivative of f prime of x equals minus m divided by x to the m plus 1.
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Now, let's see if we can see why this is true.
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F of x is 1 over x to the m power. Then the reciprocal rule tells us that f prime of x
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is minus 1 over x to the 2 m power times the derivative of x to the m, which is m times
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x to the m minus 1. Notice here, we're using the power rule that we're already familiar
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with. Now, simplifying this, we obtain minus m times x to the minus m minus 1 power, which
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is the same as minus m divided by x to the m plus 1 power.
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Now, look at the next to the last form here. The derivative of 1 over x to the m is minus
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m times x to the minus m minus 1, which looks very much like the power rule that we're
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already familiar with. Now, in fact, if we express our function as x to the minus m power,
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then what this tells us is that the very same power rule works here that works for positive
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exponents. F prime of x is minus m x to the minus m minus 1 power, which is the same
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basic rule that says multiply by the exponent and subtract one from the exponent. So what
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we now have is a more general power rule. If n is any integer, then the function f of
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x equals x to the n has derivative of f prime of x equal to n times x to the n minus 1.
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Now, for instance, let's look at f of x equals 1 over x cubed. Using the reciprocal rule,
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we have minus 1 over x to the sixth times 3 x squared, which reduces to minus 3 over x
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to the fourth. Now, we could write the same function as x to the minus 3 power, and simply
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use the power rule to write f prime of x equals minus 3 times x to the minus 4 power, which
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of course gives us the same result minus 3 divided by x to the fourth.
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The quotient rule, the derivative of a quotient f of x divided by g of x is equal to f prime
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of x times g of x minus f of x times g prime of x all divided by g of x squared. In other
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words, the derivative of the numerator times the denominator minus the numerator times
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the derivative of the denominator all divided by the denominator squared. Notice that the
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numerator here looks very similar to the product rule. Only instead of adding, we're subtracting,
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so therefore it's very important to get the order of the terms correct in the numerator.
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Always just remember that you start off with the derivative of the numerator f prime of
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x. Now, let's have a look at why this works. Let
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q of x equal f of x divided by g of x. Now, this could be written as product of f of x
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times 1 over g of x, so we can differentiate this with the product rule. q prime of x
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will be the derivative of f times 1 over g of x plus f times the derivative of 1 over g
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of x, which using the reciprocal rule is minus g prime divided by g squared. Now, we'll
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get a common denominator of g of x squared and combine terms, and there is our quotient
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rule, f prime times g minus f times g prime divided by g squared. Example, let f of x equal
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2x minus 1 divided by x plus 1, find and simplify f prime of x. We'll use the quotient rule.
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f prime of x is equal to a fraction whose denominator is the square of the denominator
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of f, and the top of the fraction is the derivative of 2x minus 1 times x plus 1 minus 2x minus
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1 times the derivative of x plus 1, which is just 1. So now, we multiply out the factors
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in the two terms in the numerator and combine terms and finally end up with 3 divided
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by x plus 1 squared. Another example, let f of x equal x cube divided by x squared
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plus 1, find and simplify f prime. Again, we use the quotient rule, f prime of x is equal
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to a quotient whose denominator is x squared plus 1 squared, and the numerator will be the
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derivative of x cubed times x squared plus 1 minus x cubed times the derivative of x squared
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plus 1. So now, we multiply out the factors in each term in the numerator and combine terms
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and we obtain x of the fourth plus 3x squared, all divided by the square of x squared plus
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1. Another example, let f of x equal the square root of x divided by x plus 1,
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again, we use the quotient rule. f prime of x will be a fraction whose denominator is
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x plus 1 squared, and we write the numerator by taking the derivative of the square root
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of x, multiplying by x plus 1, and subtracting the square root of x times the derivative
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of x plus 1. Now, to simplify this fraction, we multiply the top and bottom by 2 times
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the square root of x. This gives us a denominator that's x plus 1 squared times 2 times the
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square root of x, and the first term in the numerator becomes x plus 1, and the second
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term becomes 2 times x to the first power, and a little more simplification gives us
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1 minus x in the numerator. Let's look at the graph of this function and think about
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how it relates to the derivative. Notice that the numerator of the derivative is 1 minus
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x. That tells us that f prime of x is 0 when x is 1, moreover, the denominator is never
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negative, and so f prime of x is positive when x is between 0 and 1, and it's negative
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when x is greater than 1, and that is exactly what we see in the slope of the graph here.
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Notice that to the left of 1, the slope is positive, to the right of 1, the slope is negative.
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Summary. We've learned the following rules for differentiation. 1. The derivative of a
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sum is the sum of the derivatives. The derivative of a constant times f is the constant times
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f prime of f. The derivative of a product is the derivative of the first factor times
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the second plus the first factor times the derivative of the second. We have the reciprocal
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rule. The derivative of 1 over g is minus g prime divided by g squared. The quotient
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rule, the derivative of f divided by g is f prime times g minus f times g prime divided
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by g squared. The derivative of the numerator times the denominator minus the numerator
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times the derivative of the denominator all over the denominator squared. Also, any constant
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function f of x equals c has derivative 0. The identity function f of x equals x has
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the derivative of f prime of x equals 1. If n is any integer, then f of x equals x to
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the n has derivative of f prime of x equals n times x to the n minus 1. This is our power
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rule. And finally, the square root function f of x equals the square root of x has
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derivative of f prime of x equals 1 over 2 times the square root of x.