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

##### 1 Answer

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 #