Question

Let the function f be define as f(x)= cos(3arcsinx). Determine the value of f^(100)(0).​

Answer 1

Given:

$f(x)=\cos \left(3 \sin ^{-1} x\right)$

$\text { Take } \sin ^{-1} x=\theta \quad ; \quad f(x)=\cos (3 \theta)$

$\cos (3 \theta)=4 \cos ^{3} \theta-3 \cos \theta \quad \text { by property }$

$\therefore \cos \left(3 \sin ^{-1} x\right)=4 \cos ^{3}\left(\sin ^{-1} x\right)-3 \cos \left(\sin ^{-1} x\right)$

$\text { Taking } \sin ^{-1} x=\theta \quad \Rightarrow \sin \theta=x$

$\cos \theta=\sqrt{1-x^{2}}$

$\begin{array}{l}\because \sin ^{-1} x=\theta \\f(x)=4 \cos ^{3}(\theta)-3 \cos \theta\end{array}$

$\begin{aligned}f(x)=[&\left.4\left(\sqrt{1-x^{2}}\right)^{3}-3 \sqrt{1-x^{2}}\right] \\\therefore f^{\prime}(x) &=\left[4\left(1-\sqrt{1-x^{2}}\right)^{3}-3 \sqrt{1-x^{2}}\right]^{100} \\&=4-3 \\&=1\end{aligned}$

The answer is "1".