untchick04tm

2021-12-10

Differentiate the following function using product rule.
$f\left(x\right)=\left(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6\right)\left(2{x}^{2}-2x+4\right)$

xandir307dc

Step 1
For functions given as product of functions f(x)g(x) the derivative using product rule is given as (f(x)g(x))'=f'(x)g(x)+f(x)g'(x). One property of derivatives that will be used is (cf(x))'=cf'(x). Here c is a constant.
Another property of derivatives that will be used is $\frac{d}{dx}{x}^{n}=n{x}^{n-1}$. This is for $n\ne 0$.
Step 2
The given function is $f\left(x\right)=\left(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6\right)\left(2{x}^{2}-2x+4\right)$. Calculate the derivative using the product rule and other properties.
$f\left(x\right)=\left(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6\right)\left(2{x}^{2}-2x+4\right)$
${f}^{\prime }\left(x\right)={\left(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6\right)}^{\prime }\left(2{x}^{2}-2x+4\right)+\left(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6\right){\left(2{x}^{2}-2x+4\right)}^{\prime }$
$=\left(\frac{9{x}^{2}}{4}-4x+\frac{1}{4}\right)\left(2{x}^{2}-2x+4\right)+\left(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6\right)\left(4x-2\right)$
Hence, the derivative of given function is $\left(\frac{9{x}^{2}}{4}-4x+\frac{1}{4}\right)\left(2{x}^{2}-2x+4\right)+\left(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6\right)\left(4x-2\right)$

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