# Trouble finding the derivative of $\frac{4}{\sqrt{1-x}}$

I've been trying to figure out how to differentiate this expression, apparently I don't know my differentiation rules as much as I thought. I've been trying to use Wolfram Alpha as a guide but I'm at a loss.

I need to differentiate $$\frac{4}{\sqrt{1-x}}$$

I first pull out the four so the problem becomes: $$4 * \frac{1}{\sqrt{1-x}}$$

I'm not sure what to do next, do I use the quotient rule? Wolfram alpha is giving me some crazy answers, I would appreciate it if someone could walk me through this step by step. Thanks.

• Wolfram Alpha is useful, but do not use it for differentiation, the process needs to be internalzed and become automatic. – André Nicolas Jul 7 '14 at 6:40

## 4 Answers

A bit more detail than you asked for, but I hope it makes the problem clear:

When you differentiate $(f(x))^r$ you end up with $rf(x)^{r-1}f'(x)$. To see why, work through the following:

$$y=f(x)^r\\ \ln(y)=r\ln(f(x))$$

Now differentiate both sides:

$$(1/y)\frac{dy}{dx}=\frac{r}{f(x)}f'(x)$$

So $\frac{dy}{dx}=\frac{ry}{f(x)}f'(x)=rf'(x)f(x)^{r-1}$ when you substitute the expression for $y$ back in.

So your question comes down to : $4\frac{d}{dx}(1-x)^{-1/2}=4(-1/2)(1-x)^{-3/2}(-1)=\frac{2}{(1-x)^{3/2}}$.

In this case (like many others), despite of you are working with a quotient, the quotient rule is not needed because you can rewrite your function in a convenient way as you can see below.

\begin{align*}\frac{d}{dx}\left[\frac{4}{\sqrt{1-x}}\right]&=4\frac{d}{dx}\left[\frac{1}{\sqrt{1-x}}\right]&\text{ (basic rule)}\\ \\ &=4\frac{d}{dx}\left[\frac{1}{(1-x)^{1/2}}\right]&\text{ (rewrite because it's convenient)}\\ &=4\frac{d}{dx}\left[(1-x)^{-1/2}\right]&\text{ (rewrite again)}\\ \\ &=4\left(-\frac{1}{2}(1-x)^{-3/2}\right)\cdot\frac{d}{dx}[1-x]&\text{ (chain rule)}\\ \\ &=-\frac{2}{(1-x)^{3/2}}\cdot(-1)&\text{ (rewrite + baisc rule)}\\ \\ &=\frac{2}{\sqrt{(1-x)^3}}&\text{ (rewrite)}\\ \end{align*}

A general hint: $\sqrt{f(x)} = f(x)^{\frac{1}{2}}$ and behave it like $x^r$.

• I know that. Maybe I'm just tired and it will be clear in the morning, but could you provide further details? – Irresponsible Newb Jul 7 '14 at 4:28

Let $f(x) = 4(1-x)^{-\frac{1}{2}}$.

Then we have $f'(x) = \frac{df(x)}{dx} = 4 \times (-\frac{1}{2}) \times (-1) \times (1-x)^{-\frac{3}{2}} = 2(1-x)^{-\frac{3}{2}}$.

And we are done.