0
$\begingroup$

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 $

$\endgroup$
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

1
$\begingroup$

I will use primes to shorten the notation.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.