# Thread: Differentiating funtion of a function

1. ## Differentiating funtion of a function

I'm differentiating $\displaystyle \frac{\sqrt{1-x^2}}{1-x}$ and coming up with the answer $\displaystyle \frac{1-x}{(1-x^2)(1-x)^\frac{1}{2}}$ and an online differentiator I've run it through gives the same answer. However, the book I'm working from gives the answer as $\displaystyle \frac{1}{(1-x)(1-x^2)^\frac{1}{2}}$.

Which of these is correct? Or are they both the same and I'm just missing a way of simplifying my answer?

2. Maybe they are the same function? You need to use the quotient rule here.

3. My working goes like this:

Let $\displaystyle u=\sqrt{1-x^2}$ and $\displaystyle v=1-x$

Therefore $\displaystyle \frac{du}{dx}=\frac{1}{2}(1-x^2)^\frac{-1}{2}\times -2x=\frac{-x}{(1-x^2)^\frac{1}{2}}$ and $\displaystyle \frac{dv}{dx}=-1$

Therefore

$\displaystyle \frac{dy}{dx}=\frac{(1-x)\frac{-x}{(1-x^2)^\frac{1}{2}}+(1-x^2)^\frac{1}{2}}{(1-x)^2}$

$\displaystyle =\frac{\frac{-x(1-x)}{(1-x^2)^\frac{1}{2}}+(1-x^2)^\frac{1}{2}}{(1-x^2)^\frac{1}{2}}$

$\displaystyle =\frac{-x(1-x)+(1-x^2)}{(1-x)^2(1-x^2)^\frac{1}{2}}$

$\displaystyle =\frac{1-x}{(1-x)^2(1-x^2)^\frac{1}{2}}$

4. you have a common factor of 1-x in the numerator and denominator, so they cancel. your original posted answer is incorrect, your second answer with the work shown IS correct, and equal to the answer in your book.