# Find the derivative of f(x)=(x^2+8)(8x-1) by first expanding the polynomials. Enter the fully simplified expression for f(x) after expanding the polynomials

Find the derivative of $f\left(x\right)=\left({x}^{2}+8\right)\left(8x-1\right)$ by first expanding the polynomials.
Enter the fully simplified expression for f (x) after expanding the polynomials
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Emaidedip6g
$f\left(x\right)=\left({x}^{2}+8\right)\left(8x-1\right)$
find derivative=?
We use the substitutions $u={x}^{2}+8$ and v=8x-1
Now, use product rule
$\frac{d\left(uv\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\phantom{\rule{0ex}{0ex}}=\left({x}^{2}+8\right)\frac{d}{dx}\left(8x-1\right)+\left(8x-1\right)\frac{d}{dx}\left({x}^{2}+8\right)$
Now, we differentiate
$=\left({x}^{2}+8\right)×\left(8\right)+\left(8x-1\right)×\left(2x\right)\phantom{\rule{0ex}{0ex}}=8{x}^{2}+64+16{x}^{2}-2x\phantom{\rule{0ex}{0ex}}=24{x}^{2}-2x+64\phantom{\rule{0ex}{0ex}}f\left(x\right)=24{x}^{2}-2x+64$
Now, find f'(x)=?
$f\left(x\right)=24{x}^{2}-2x+64$
differentiate with respect to x
${f}^{\prime }\left(x\right)=48x-2$