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If $f(x)=\cos ^{-1}\left[\frac{1}{\sqrt{13}}(2 \cos x-3 \sin x)\right]$. Then, $f^{\prime}(0.5)$ is equal to
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Given, $f(x)=\cos ^{-1}\left\{\frac{1}{\sqrt{13}}(2 \cos x-3 \sin x)\right\}$
$\Rightarrow f(x)=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos x-\frac{3}{\sqrt{13}} \sin x\right\}$
$\Rightarrow \quad f(x)=\cos ^{-1}\{\cos \alpha \cdot \cos x-\sin \alpha \cdot \sin x\}$
$\begin{aligned}
&=\cos ^{-1}\{\cos (x+\alpha)\} \\
&=x+\alpha
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
f^{\prime}(x) &=1=\text { constant value } \\
\therefore \quad f^{\prime}(0.5) &=1
\end{aligned}$
$\Rightarrow f(x)=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos x-\frac{3}{\sqrt{13}} \sin x\right\}$
$\Rightarrow \quad f(x)=\cos ^{-1}\{\cos \alpha \cdot \cos x-\sin \alpha \cdot \sin x\}$
$\begin{aligned}
&=\cos ^{-1}\{\cos (x+\alpha)\} \\
&=x+\alpha
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
f^{\prime}(x) &=1=\text { constant value } \\
\therefore \quad f^{\prime}(0.5) &=1
\end{aligned}$
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