How do you differentiate #f(x)=(x^2)/(e^(x^-3))# using the quotient rule?

Answer 1

# f'(x) = ( x(2-x) ) / (e^(x-3)) #

The quotient rule is #d/dx(u/v)=(v(du)/dx - u(dv)/dx)/v^2#

So, # f(x)=x^2/e^(x-3) # # :. f'(x) = ( e^(x-3) d/dx(x^2)-x^2d/dx(e^(x-3)) ) / (e^(x-3))^2 # # :. f'(x) = ( e^(x-3) (2x)-x^2(e^(x-3)) ) / (e^(x-3))^2 # # :. f'(x) = ( e^(x-3) (2x-x^2) ) / (e^(x-3))^2 # # :. f'(x) = ( (2x-x^2) ) / (e^(x-3)) # # :. f'(x) = ( x(2-x) ) / (e^(x-3)) #

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Answer 2

Ezekiel Alexander

To differentiate f(x) = (x^2)/(e^(x^-3)) using the quotient rule, we use the formula:

f'(x) = [g(x)*f'(x) - f(x)*g'(x)] / [g(x)]^2

Where f(x) = x^2 and g(x) = e^(x^-3).

Now, let's find the derivatives: f'(x) = 2x g'(x) = -3e^(x^-3)/(x^4)

Now, apply the quotient rule:

f'(x) = [(e^(x^-3)2x) - ((x^2)(-3e^(x^-3)/(x^4)))] / [e^(x^-3)]^2 = [(2xe^(x^-3)) + (3x^2e^(x^-3)/(x^4))] / e^(2(x^-3)) = [(2xe^(x^-3)) + (3x^2e^(x^-3-x^4))] / e^(2(x^-3))

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.