find $f(x)$ for the function.\n$f(x)=8x(18x^{4}+\frac{16}{x + 1})$ \n$f(x)=square$

Explanation:

Step1: Expand the function

$f(x)=8x\times18x^{4}+8x\times\frac{16}{x + 1}=144x^{5}+\frac{128x}{x + 1}$

Step2: Differentiate the first - term

The derivative of $y = 144x^{5}$ using the power rule $(x^n)'=nx^{n - 1}$ is $y'=144\times5x^{4}=720x^{4}$

Step3: Differentiate the second - term

Use the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = 128x$, $u'=128$, $v=x + 1$, $v'=1$. Then $\left(\frac{128x}{x + 1}\right)'=\frac{128(x + 1)-128x\times1}{(x + 1)^{2}}=\frac{128x+128 - 128x}{(x + 1)^{2}}=\frac{128}{(x + 1)^{2}}$

Step4: Find the derivative of $f(x)$

$f'(x)=720x^{4}+\frac{128}{(x + 1)^{2}}$

Answer:

$720x^{4}+\frac{128}{(x + 1)^{2}}$