Question

If $y={\cos }^{-1}\left[\frac{3\cos x-4\sin x}{5}\right]$ then $\frac{dy}{dx}$ equals to

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

The correct option is B

$1$

Explanation for the correct option.

Step 1. Simplify the given equation.

Let,, $\cos \theta =\frac{3}{5}$, then $\sin \theta$ is given as:

$\begin{array}{rcl}\sin \theta & =& \sqrt{1-{\cos }^2\theta }\\ & =& \sqrt{1-{\left(\frac{3}{5}\right)}^2}\\ & =& \sqrt{1-\frac{9}{25}}\\ & =& \sqrt{\frac{16}{25}}\\ & =& \frac{4}{5}\end{array}$

So, using the values of $\sin \theta$ and $\cos \theta$ the equation can be modified as:

$\begin{array}{rcl}y& =& {\cos }^{-1}\left[\frac{3\cos x-4\sin x}{5}\right]\\ & =& {\cos }^{-1}\left[\frac{3}{5}\cos x-\frac{4}{5}\sin x\right]\\ & =& {\cos }^{-1}\left[\cos \theta \cos x-\sin \theta \sin x\right]\\ & =& {\cos }^{-1}\left(\cos \left(\theta +x\right)\right)\left[\cos (A+B)=\cos A\cos B-\sin A\sin B\right]\\ & =& \theta +x\end{array}$

Step 2. Find the value of $\frac{dy}{dx}$.

Differentiate $y=\theta +x$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}(\theta +x)\\ & =& 0+1\\ & =& 1\end{array}$

Hence, the correct option is B.