Jerimiah Boone

2022-03-15

What is the instantaneous rate of change of $f\left(x\right)=({x}^{2}-2){e}^{{x}^{2}-3}$ at x=2?

suboctavaja5

Beginner2022-03-16Added 6 answers

This is the same as evaluating f'(2)

Differentiate using the Product rule

If f(x) = g(x)*h(x) then f'(x) = g(x)*h'(x) + h(x)*g'(x)

here:$g\left(x\right)=({x}^{2}-2)\Rightarrow {g}^{\prime}\left(x\right)=2x$

and$h\left(x\right)={e}^{{x}^{2}-3}\Rightarrow {h}^{\prime}\left(x\right)={e}^{{x}^{2}-3}\cdot \frac{d}{dx}({x}^{2}-3)=2x{e}^{{x}^{2}-3}$

now substitute these values into f'(x)

${f}^{\prime}\left(x\right)=({x}^{2}-2)\cdot 2x\cdot {e}^{{x}^{2}-3}+{e}^{{x}^{2}-3}\cdot 2x$

factorising to obtain

${f}^{\prime}\left(x\right)=2x\cdot {e}^{{x}^{2}-3}[{x}^{2}-2+1]$

$\Rightarrow {f}^{\prime}\left(2\right)=4{e}^{1}\left(3\right)=12e\approx 32.62$

Differentiate using the Product rule

If f(x) = g(x)*h(x) then f'(x) = g(x)*h'(x) + h(x)*g'(x)

here:

and

now substitute these values into f'(x)

factorising to obtain

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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