Differentiate the following function $f(x)=\sqrt{\frac{1-x}{1+x}}$

Question

$f(x)=\sqrt{\frac{1-x}{1+x}}$, find $f'(x)$.

I know I have to use the quotient rule but, but I got confused by the square root?

Answer 1

Hint: Use the power and the chain rule: $$y'=\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{-1/2}\left(\frac{-1(1+x)-(1-x)}{(1+x)^2}\right)$$

Answer 2

First, you need to apply the rule for composition: $(g(h(x)))'=g'(h(x))\cdot h'(x)$. You obtain $\frac{1}{2\sqrt{\frac{1-x}{1+x}}}\cdot (\frac{1-x}{1+x})'$. Try to finish from here.

Answer 3

$$ \frac{d}{dx}\left( \sqrt{\frac{f(x)}{g(x)}} \right)= \frac{\frac{d}{dx} \left(\frac{f(x)}{g(x)} \right)}{2\sqrt{\frac{f(x)}{g(x)}}}= \frac{\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} }{2\sqrt{\frac{f(x)}{g(x)}}}= \frac{f'(x)g(x)-f(x)g'(x)}{2g^2(x)\sqrt{\frac{f(x)}{g(x)}} } $$ I hope that this can help

Answer 4

A couple of ways...

1) Square both sides and implicit differentiation:

Let $y = \sqrt{\frac{1-x}{1+x}}$, then $y^2 = \frac{1-x}{1+x}$.

Hence $$2y \frac{dy}{dx} = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2}$$ $$\Rightarrow \frac{dy}{dx} = \frac{1}{2y} \times \frac{-2}{(1+x)^2} = -\frac{1}{(1+x)^2}\sqrt{\frac{1+x}{1-x}} $$

Haven't spelt the last step out explicitly, but work through it and replace $y$ and you'll get to the result...

2) Chain rule:

$$\frac{dy}{dx} = \frac{1}{2} \left(\frac{1-x}{1+x}\right)^{-\frac{1}{2}} \times \frac{d}{dx} \left(\frac{1-x}{1+x}\right)$$ $$\Rightarrow\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1+x}{1-x}} \times \frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2}$$ and this simplifies to the exact same result, which was $$\frac{dy}{dx}= -\frac{1}{(1+x)^2}\sqrt{\frac{1+x}{1-x}} $$

Answer 5

You can also use logarithmic differentiation $$f=\sqrt{\frac{1-x}{1+x}}\implies \log(f)=\frac 12 \log(1-x)-\frac 12 \log(1+x)$$

$$\frac{f'}f=-\frac 12 \frac 1 {1-x}-\frac 12 \frac 1 {1+x}=-\frac 1{(1+x)(1-x)}$$ Now $$f'=f \times \frac{f'}f$$ and simplify.

Answer 6

So you want to differentiate $f(x)= \left(\frac{1- x}{1+ x}\right)^{1/2}$? Your title implies that you want to differentiate some other function with respect to f!

Using the "chain rule" the derivative is $\frac{1}{2}\left(\frac{1- x}{1+ x}\right)^{-1/2}$ times the derivative of $\frac{1- x}{1+ x}$. Using the "quotient rule", $\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$, the derivative of the fraction is $\frac{-1(1+ x)- 1(1- x)}{(1+ x)^2}= \frac{-2}{(1+ x)^2}$.

So $f'= \frac{-2}{(1+ x)^2}\left(\frac{1- x}{1+ x}\right)^{-1/2}$.