Find the derivative using the chain rule and the quotient rule

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$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.

Here is my work:

$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$

$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$

I know the final simplified answer to be:

$${4x^3\over (x+1)^5}$$

How do I get to the final answer from my last step? Or have I done something wrong in my own work?

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6 Answers

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You calculation of $f'(x)$ is correct, though it can be simplified. To do this, we factor out $4x^3(x+1)^3\require{cancel}$ in the numerator.

$$\begin{align}f'(x) &= \frac{4x^3(x+1)^4-4x^4(x+1)^3}{(x+1)^8}\\\\ &= \frac{\color{blue}{4x^3(x+1)^3}\Big((x+1) - x\Big)}{(x+1)^8}\\\\ &= \frac {(4x^3\cancel{(x+1)^3})(1)}{\cancel{(x+1)^8}^5}\\ \\ &= \frac{4x^3}{(x+1)^5} \end{align}$$

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it is easier to use logarithmic differentiation on this problem. here is how you do this. $$y = \left(\frac x{x+1}\right)^4 \to \ln y = 4 \ln x - 4 \ln x + 1\to \frac{dy}{y} =4\left(\frac{dx}{x} - \frac{dx}{x+1} \right) = \frac{4dx}{x(x+1)} $$ multiplying through by $y$ gives you $$\frac{dy}{dx} = \frac{4x^3}{(x+1)^5} $$

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Apparently you complicated things too much.

Starting from $$f(x) = \left(\frac{x}{x+1}\right)^4$$ with the chain rule we get first \begin{align} f'(x) &= 4\left(\frac{x}{x+1}\right)^3\cdot \left(\frac{x}{x+1}\right)^\prime\notag\\ &= 4\left(\frac{x}{x+1}\right)^3\cdot \frac{x'(x+1)-x(x+1)'}{(x+1)^2}\notag\\ &= 4\frac{x^3}{(x+1)^3}\cdot \frac{(x+1)-x}{(x+1)^2}\notag\\ &= \frac{4x^3}{(x+1)^3}\cdot \frac 1{(x+1)^2}\notag\\ &= \frac{4x^3}{(x+1)^5}\notag \end{align}

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I usually try to avoid the quotient rule if possible.

$$ f(x) = \left(\frac{x}{x+1}\right)^4 = \left(\frac{x+1-1}{x+1}\right)^4 = \left(\frac{x+1}{x+1} + \frac{-1}{x+1}\right)^4 = \left(1 - \frac{1}{x+1}\right)^4 $$

Applying the chain rule yields

$$ f'(x) = 4 \cdot \left(1-\frac{1}{x+1}\right)^3 \cdot \frac{1}{(x+1)^2} = 4 \cdot \left(\frac{x}{x+1}\right)^3 \cdot \frac{1}{(x+1)^2} = 4 \cdot \frac{x^3}{(x+1)^3} \cdot \frac{1}{(x+1)^2} = \frac{4x^3}{(x+1)^5} $$

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Not wrong, only written in a different form: you can simplify much of it.

It's much better exploiting the chain rule: if you call $$ g(x)=\frac{x}{x+1}, $$ then $f(x)=(g(x))^4$ and so $$ f'(x)=4(g(x))^3g'(x)= 4\left(\frac{x}{x+1}\right)^{\!3}\frac{1(x+1)-x\cdot 1}{(x+1)^2}= \frac{4x^3}{(x+1)^5} $$

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First use the chain rule $$(4*x^3)/(x+1)^3$$........() Now derivate the inner function finding using quotient rule $$((X+1)-x)/(x+1)^{2}=> 1/(x+1)^{2}$$.........(${*}{*}$) Now multiply them both () &. (**) to get &$${4x^{3}}/{(x+1)^{5}}$$

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