What is the instantaneous rate of change of f(x)=(x^2-2)e^(x^2-3) at x=2 ?

1 Answer
Jim G.
Mar 24, 2016

12e ≈ 32.62

Explanation:

This is the same as evaluating f'(2)

Differentiate using the color(blue)" Product rule "

If f(x) = g(x).h(x) then f'(x) = g(x).h'(x) + h(x).g'(x)
" --------------------------------------------------"
here: g(x) =(x^2 - 2 ) rArr g'(x) = 2x

and h(x) = e^(x^2-3) rArr h'(x) = e^(x^2-3). d/dx(x^2-3)=2xe^(x^2-3)
"------------------------------------------------------"
now substitute these values into f'(x)

f'(x) = (x^2-2).2x.e^(x^2-3) + e^(x^2-3).2x

factorising to obtain

f'(x) = 2x.e^(x^2-3)[x^2-2+1]

rArr f'(2) = 4e^1(3) = 12e ≈ 32.62

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