A) Compute the derivative using the quotient rule (without simplifying all the way) of the...


A) Compute the derivative using the quotient rule (without simplifying all the way) of the function {eq}\; h(x) = \frac{x(x + 1)^3}{(3x - 1)^2} {/eq}.

B) Repeat this process, but using logarithmic differentiation instead.

C) Check that {eq}{h}'(1) {/eq} is the same for both methods.

Logarithmic Differentiation

In some problems where other derivative rules may be used - but may get complicated - logarithmic differentiation often makes things easier. Logarithmic differentiation is a technique in which before taking a derivative, a logarithm is taken of both sides of the equation. Then, using properties of logarithms, the problem can be rewritten in a simpler manner. The derivative can then be evaluated using implicit differentiation and then solved for.

Answer and Explanation:

For the function {eq}h(x) = \dfrac{x(x + 1)^3}{(3x - 1)^2} {/eq}

A) Using the quotient rule (without simplifying) we have

{eq}h'(x) = \dfrac{(3x-1)^2(x(3(x+1)^2) + (x+1)^3) - x(x+1)^3(2(3x-1)(3))}{(3x-1)^4}\\ h'(x) = \dfrac{3x(x+1)^2(3x-1)^2 + (x+1)^3(3x-1)^2 - 6x(x+1)^3(3x-1)}{(3x-1)^4} {/eq}

B) Using logarithmic differentiation instead, begin by taking a logarithm of both sides of the equation and using logarithmic properties to simplify

{eq}\ln(h(x)) = \ln\left(\dfrac{x(x + 1)^3}{(3x - 1)^2}\right)\\ \ln(h(x)) = \ln\left(x(x+1)^3\right) - \ln\left((3x-1)^2\right)\\ \ln(h(x)) = \ln(x) + 3\ln(x+1) - 2\ln(3x-1) {/eq}

Then take the derivative using implicit differentiation

{eq}\dfrac{1}{h(x)}h'(x) = \dfrac{1}{x} + \dfrac{3}{x+1} - \dfrac{6}{3x-1}\\ h'(x) = h(x)\left(\dfrac{1}{x} + \dfrac{3}{x+1} - \dfrac{6}{3x-1}\right)\\ h'(x) = \dfrac{x(x + 1)^3}{(3x - 1)^2}\left(\dfrac{1}{x} + \dfrac{3}{x+1} - \dfrac{6}{3x-1}\right)\\ {/eq}

C) Substituting in x=1 into our results for A and B, we have

{eq}h'(1) = \dfrac{3(1)(1+1)^2(3(1)-1)^2 + (1+1)^3(3(1)-1)^2 - 6(1)(1+1)^3(3(1)-1)}{(3(1)-1)^4}\\ h'(1) = \dfrac{3(2)^2(2)^2 + (2)^3(2)^2 - 6(2)^3(2)}{2^4}\\ h'(1) = -\dfrac{16}{16}\\ h'(1) = -1 {/eq}


{eq}h'(1) = \dfrac{1(1 + 1)^3}{(3(1) - 1)^2}\left(\dfrac{1}{1} + \dfrac{3}{1+1} - \dfrac{6}{3(1)-1}\right)\\ h'(1) = \dfrac{2^3}{2^2}\left(1+\dfrac{3}{2} - \dfrac{6}{2}\right)\\ h'(1) = 2\left(-\dfrac{1}{2}\right)\\ h'(1) = -1 {/eq}

Learn more about this topic:

When to Use the Quotient Rule for Differentiation

from Math 104: Calculus

Chapter 8 / Lesson 8

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