Answer to Question #151173 in Calculus for Angelo

Question

Differentiate the Inverse Trigonometric functions:

1. Find 𝑓′′(𝜋) if 𝑓(𝑥) = 𝑎𝑟𝑐𝑠𝑖𝑛(𝑐𝑜𝑠𝑥).

2. If 𝑐𝑜𝑡^−1(𝑥𝑦) − 𝑡𝑎𝑛^−1(𝑦/𝑥) = 0, find 𝑑𝑦/𝑑𝑥.

3. ℎ(𝑡) = 𝑎𝑟𝑐𝑐𝑜𝑠 (𝑡−1/𝑡+1).

Answer 1

1) Here $f(x)=arcsin(cosx)$

Therefore $f'(x)=\frac{1}{\sqrt{1-cos^2x}}.\frac{d}{dx}(cosx)$

$=\frac{1}{\sqrt {sin^2x}}.(-sinx)$

$=\frac{(-sinx)}{sinx}$

$=-1$

$\therefore f''(x)=0$

Hence, $f''(\pi)=0$

2)Given that , $Cot^{-1}(xy)-tan^{-1}(\frac {y}{x})=0$ .........$(1)$

Now differentiating both side of (1) with respect to $x$ ,we get

$\implies \frac{(-1)}{1+x^2y^2}.\frac{d}{dx}(xy)-\frac{1}{1+(\frac{y}{x})^2}.\frac{d}{dx}(\frac{y}{x})=0$

$\implies$ $\frac{1}{1+x^2y^2}.(y+x.\frac{dy}{dx})+(\frac{x^2}{x^2+y^2}).(\frac{x\frac{dy}{dx}-y}{x^2})=0$

$\implies$ $\frac{y}{1+x^2y^2}+\frac{x\frac{dy}{dx}}{1+x^2y^2}+\frac{x\frac{dy}{dx}}{x^2+y^2}-\frac{y}{x^2+y^2}=0$

$\implies y.(\frac{1}{1+x^2y^2}-\frac{1}{x^2+y^2})+x\frac{dy}{dx}.(\frac{1}{1+x^2y^2}+\frac{1}{x^2+y^2})=0$

$\implies y.[\frac{x^2+y^2-x^2y^2-1}{(1+x^2y^2)(x^2+y^2)}]+x\frac{dy}{dx}.[\frac{x^2+y^2+x^2y^2+1}{(1+x^2y^2)(x^2+y^2)}]=0$

$\implies \frac {dy}{dx}=-(\frac {y}{x}).[\frac {x^2+y^2-x^2y^2-1}{x^2+y^2+x^2y^2+1}]$

3) Given ,$h(t)=arccos(\frac {t-1}{t+1})$

Then $h'(t)=\frac{-1}{\sqrt{1-(\frac {t-1}{t+1})^2}}.\frac{d}{dt}(\frac{t-1}{t+1})$

$=\frac{-1}{\sqrt{1-(\frac {t-1}{t+1})^2}}.[\frac {(t+1). \frac{d}{dt}(t-1)-(t-1).\frac{d}{dt}(t+1)}{(t+1)^2}]$

$=\frac {-(t+1)}{\sqrt{(t+1)^2-(t-1)^2}}. [\frac {(t+1). 1-(t-1).1}{(t+1)^2}]$

$=\frac {-(t+1)}{\sqrt{4t}}. [\frac {2}{(t+1)^2}]$

$=\frac{-(t+1)}{2\sqrt{t}}.[\frac{2}{(t+1)^2}]$

$=\frac{-1}{\sqrt{t}.(t+1)}$