Answer
$f'(x)=\dfrac{x^{2}+2x-7}{(x+1)^{2}}$
$f''(x)=\dfrac{16}{(x+1)^{3}}$
Work Step by Step
$f(x)=\dfrac{x^{2}-7x}{x+1}$
Use the quotient rule to evaluate the first derivative of the function given:
$f'(x)=\dfrac{(x+1)(x^{2}-7x)'-(x^{2}-7x)(x+1)'}{(x+1)^{2}}=...$
Evaluate the derivatives indicated in the numerator:
$...=\dfrac{(x+1)(2x-7)-(x^{2}-7x)(1)}{(x+1)^{2}}=...$
Evaluate the products indicated in the numerator and simplify:
$...=\dfrac{2x^{2}-5x-7-x^{2}+7x}{(x+1)^{2}}=\dfrac{x^{2}+2x-7}{(x+1)^{2}}$
The expression above is the first derivative.
Apply the quotient rule again to evaluate the second derivative:
$f''(x)=\dfrac{(x+1)^{2}(x^{2}+2x-7)'-(x^{2}+2x-7)[(x+1)^{2}]'}{(x+1)^{4}}=...$
$...=\dfrac{(x+1)^{2}(2x+2)-2(x^{2}+2x-7)(x+1)}{(x+1)^{4}}=...$
Simplify:
$...=\dfrac{2x^{3}+6x^{2}+6x+2-2x^{3}-6x^{2}+10x+14}{(x+1)^{4}}=...$
$...=\dfrac{16x+16}{(x+1)^{4}}=\dfrac{16(x+1)}{(x+1)^{4}}=\dfrac{16}{(x+1)^{3}}$
The expression above is the second derivative
