2

I am trying to differentiate the below with respect to c: $\left(\frac{a-b}{c-b}\right)^d$, however I get an answer different to what Mathematica (and other sources) is telling me, which is $-\frac{(a-b)\left(\frac{a-b}{c-b}\right)^{d-1}d}{(c-b)^2}$

The way I'm approaching it is: $\left(\frac{a-b}{c-b}\right)^d = \frac{(a-b)^d}{(c-b)^d} = (a-b)^d*(c-b)^{-d}$, then I treat $(a-b)^d$ as a constant and differentiate the $(c-b)^{-d}$ term which gives me: $(a-b)^d * -d(c-b)^{-d-1} = \frac{-d(a-b)^d}{(c-b)^{d-1}}$.

I can't work out what rule or a method I'm missing here unfortunately so I would be very grateful for any advice or pointers on that. Thanks for reading!

coconut
  • 31

1 Answers1

5

The error is in the last transformation of you work:

$$(c-b)^{-d-1} = \frac1{{(c-b)}^{d\color{red}+1}}$$

Taking this correction into account, you'll see that then your and Mathematica's answer are actually the same.

Ingix
  • 14,494