I am looking for some help on this relatively simple chain rule derivative.

I think I know where the issues are, but I cannot figure out the right steps.

Below is the problem with what I am working with so far.

Issues: I should be applying quotient rule but I forget where it should be going (I think it goes in place of A = $\frac{1}{B} $ ). Even still, I do not have two primes to satisfy the quotient rule here $\frac{B'C-C'B}{C^2} $ .

m(x) = $e ^ \frac{1}{x2+2x-2}$

$m(x) = e^A$

A = $\frac{1}{B} $

B = $x^2+2x-2 $

$\frac{dm}{dA} = e^A$

$\frac{dA}{dB} = \frac{-1}{B^2}$

$\frac{dB}{dX} = 2x+2 $

  • 1
    $\begingroup$ Looks good. Now just say $\frac {dm}{dx} = \frac{dm}{dA}\frac{dA}{dB}\frac{dB}{dx}$ complete some substitutions and you are done. $\endgroup$
    – Doug M
    Sep 3, 2016 at 2:37
  • $\begingroup$ Thanks a lot, I did not consider this way of solving! $\endgroup$
    – John Stud
    Sep 3, 2016 at 2:44

1 Answer 1


I will use primes to shorten the notation.

$$m'(x) = A' e^A = (B^{-1})' e^A = -B^{-2} B'e^A = -(2x+2)(x^2+2x-2)^{-2} e^{\frac{1}{x^2+2x+2}}$$


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