Differentiation with respect to a complex expression

[Karol]

Homework Statement

Homework Equations

$$(x-a)(x+a)=x^2-a^2$$

The Attempt at a Solution

I have to express ##~\displaystyle x^2+16=f\left( \frac{x}{x-1} \right)##
I guess it has to be ##~\displaystyle \left( \frac{x}{x-1} \right)^n-a~## or ##~\displaystyle \left( \frac{x}{x-1} \pm a \right)^n##
I tried ##~\displaystyle \left( \frac{x}{x-1}+4 \right)^2~## but no good

 

[andrewkirk]

Use substitution. Write ##y=\sqrt{x^2+16}## and ##t=x/(x-1)##. Then by the chain rule:
$$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$$
and both factors on the RHS are easily calculated (the second by expressing ##x## in terms of ##t##).

 

[Karol]

$$\frac{dy}{dx}=\frac{1}{2}(x^2+16)^{-1/2}\cdot 2x=\frac{x}{\sqrt{x^2+16}}$$
$$x=\frac{t}{t-1},~\frac{dx}{dt}=\frac{-1}{t-1}$$
$$\frac{dy}{dt}=\frac{x}{\sqrt{x^2+16}}\frac{-1}{t-1}=\frac{x}{\sqrt{x^2+16}}(1-x)$$
$$x=3~\rightarrow~\frac{dy}{dt}=\frac{3(-2)}{5}=-\frac{6}{5}$$
The answer should be ##~\displaystyle -\frac{12}{5}##

 

[Karol]

Correction:
$$x=\frac{t}{t-1},~~\frac{dx}{dt}=\frac{-1}{(t-1)^2}=-(x-1)^2$$
$$\frac{dy}{dt}=\frac{-x(x-1)^2}{\sqrt{x^2+16}}$$
$$x=3~\rightarrow~\frac{dy}{dt}=-\frac{12}{5}$$