untchick04tm

2021-12-10

Differentiate the following function using product rule.

$f\left(x\right)=(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6)(2{x}^{2}-2x+4)$

xandir307dc

Beginner2021-12-11Added 35 answers

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$f\left(x\right)=(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6)(2{x}^{2}-2x+4)$ . Calculate the derivative using the product rule and other properties.

$f\left(x\right)=(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6)(2{x}^{2}-2x+4)$

$f}^{\prime}\left(x\right)={(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6)}^{\prime}(2{x}^{2}-2x+4)+(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6){(2{x}^{2}-2x+4)}^{\prime$

$=(\frac{9{x}^{2}}{4}-4x+\frac{1}{4})(2{x}^{2}-2x+4)+(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6)(4x-2)$

Hence, the derivative of given function is$(\frac{9{x}^{2}}{4}-4x+\frac{1}{4})(2{x}^{2}-2x+4)+(\frac{3}{4}{x}^{3}-2{x}^{2}+\frac{x}{4}-6)(4x-2)$

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

Step 2

The given function is

Hence, the derivative of given function is

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