$$f(x) = \sqrt{1+\cot^2(x)}$$
How to calculate the derivative $f'(x)$? I've been looking at similar problems in my book and at examples, but I'm having a lot of trouble understanding it still. I'd appreciate it if someone could explain this in a way that would help not just answer this but other questions like it.
$\endgroup$2 Answers
$\begingroup$We have $$ 1+\cot^2x=\frac{\sin^2x}{\sin^2x} + \frac{\cos^2x}{\sin^2x}\\ =\frac{1}{\sin^2x} $$ So $f(x) =1/\sqrt{\sin^2x} =1/|\sin x|$. Can you take it from there?
$\endgroup$ 3 $\begingroup$Let $g(x) = \sqrt{x}, h(x) = 1 + x^2, k(x) = \cot x$.
Then $f(x) = g(h(k(x)))$ and $\frac {df}{dx} = \frac{dg}{dh} \frac{dh}{dk} \frac{dk}{dx}$.
$$\frac{dg}{dh} = \frac{1}{2\sqrt{h(k(x))}}$$
$$\frac{dh}{dk} = 2k(x)$$
$$\frac{dk}{dx} = -\frac{1}{\sin^2x}$$
Can you continue by yourself from here?
