Question

Finding a Derivative, find the derivative of the algebraic function.

Answer 1

We are given the function:

$f(x)=x^4(1-\frac{2}{x+1})$

To find its derivative, we can distribute x^4 first so that we won't have to apply both the product rule and the quotient rule.

$f(x)=x^4-\frac{2x^4}{x+1}$

So we'll just solve for the derivative of the first term, x^4, then add it to the derivative of the second term, 2x^4/(x+1).

$f^{\prime}(x)=g^{\prime}(x)+h^{\prime}(x)$

where g(x) = x^4 and h(x) = 2x^4/(x+1).

$\begin{gathered} g(x)=x^4 \\ g^{\prime}(x)=4x^{4-1} \\ g^{\prime}(x)=4x^3 \end{gathered}$$\begin{gathered} h(x)=\frac{2x^4}{x+1} \\ \\ h^{\prime}(x)=\frac{(x+1)(2)(4)(x^{4-1})-(2x^4)(1+0)}{(x+1)^2} \\ \\ h^{\prime}(x)=\frac{(x+1)(8x^3)-(2x^4)}{(x+1)^2} \\ \\ h^{\prime}(x)=\frac{8x^4+8x^3-2x^4}{(x+1)^2} \\ \\ h^{\prime}(x)=\frac{6x^4+8x^3}{(x+1)^2} \\ \\ h^{\prime}(x)=\frac{2x^3(3x+4)}{(x+1)^2} \end{gathered}$

So f'(x) must be:

$f^{\prime}(x)=4x^3+\frac{2x^{3}(3x+4)}{(x+1)^{2}}$