Question

If $y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$, then $\frac{dy}{dx}$is equal to

• A. $\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{-1}{\sqrt{1-x^2}}$
• C. $\frac{3}{\sqrt{1-x^2}}$
• D. $\frac{x}{\sqrt{1-x^2}}$

Answer 1

The correct option is A

$\frac{1}{\sqrt{1-x^2}}$

Explanation for the correct option:

Step 1: Simplifying the given equation.

$y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$

Put $x=\sin \theta$

$\Rightarrow \theta ={\sin }^{-1}x...\left(1\right)$

$$\begin{gathered} y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right) \\ ={\sin }^{-1}\left(\frac{5\sin \theta +12\sqrt{1-{\sin }^2\theta }}{13}\right) \\ ={\sin }^{-1}\left(\frac{5\sin \theta +12\sqrt{{\cos }^2\theta }}{13}\right)\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right] \\ ={\sin }^{-1}\left(\frac{5\sin \theta +12\cos \theta }{13}\right) \end{gathered}$$

Step 2: Finding the value of $\frac{dy}{dx}$:

Let $r\cos \alpha =5...\left(2\right)$

and $r\sin \alpha =12...\left(3\right)$

Divide $\left(3\right)$ by $\left(2\right)$, we get

$$\begin{gathered} \tan \alpha =\frac{12}{5} \\ \Rightarrow \alpha ={\tan }^{-1}\left(\frac{12}{5}\right)...\left(4\right) \end{gathered}$$

Now, Adding the square of both the equations:

$$\begin{gathered} r^2{\cos }^2\alpha +r^2{\sin }^2\alpha =5^2+{12}^2 \\ \Rightarrow r^2\left({\cos }^2\alpha +{\sin }^2\alpha \right)=169 \\ \Rightarrow r^2={13}^2 \\ \Rightarrow r=13 \end{gathered}$$

So, we have

$$\begin{gathered} y={\sin }^{-1}\left(\frac{r\cos \alpha \sin \theta +r\sin \alpha \cos \theta }{r}\right) \\ ={\sin }^{-1}\left(\sin \right(\theta +\alpha \left)\right)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{sin}\left(\theta +\alpha \right)\mathbf{=}\mathbf{sin\theta cos\alpha }\mathbf{+}\mathbf{sin\alpha cos\theta }\mathbf{]} \\ =\theta +\alpha \\ ={\sin }^{-1}x+{\tan }^{-1}\left(\frac{12}{5}\right)\left[\mathrm{from}\left(1\right)\mathrm{and}\left(4\right)\right] \end{gathered}$$

Differentiating with respect to $x$ both sides:

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{\sqrt{1-x^2}}+0\\ & =& \frac{1}{\sqrt{1-x^2}}\end{array}$

Hence, Option (A) is the correct answer.