Question

The derivative of sin^-1 (2x² - 1)  w.r.t sin^-1 x is. 
1. \(\sqrt{2}\)
2. 2
3. 2\(\sqrt{2}\)
4. 1

Answer 1

Correct Answer - Option 2 : 2

Concept:  

Parametric derivative  , \(\rm\frac{\mathrm{d}y }{\mathrm{d} z}= \frac{\frac{dy}{dx}}{\frac{dz}{dx}}\)   

\(\rm \frac{\mathrm{d} (sin^{-1}x)}{\mathrm{d} x} = \frac{1}{\sqrt{1-x^{2}}}\)

Calculation: 

Let , y = sin-1 (2x2 - 1) and z = sin-1 x 

Differentiate both the function w.r.t to x 

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} \sin^{-1}\left ( 2x^{2} -1\right )}{\mathrm{d} x}\) 

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\sqrt{1-\left ( 2x^{2} -1\right )^{2}}}\times \left ( 4x-0 \right )\) 

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2}{\sqrt{1-x^{2}}}\)           ....(i)

Similarly , 

\(\rm\frac{\mathrm{d} z}{\mathrm{d} x}= \frac{1}{\sqrt{1-x^{2}}}\)   ....(ii)

We know that , \(\rm\frac{\mathrm{d}y }{\mathrm{d} z}= \frac{\frac{dy}{dx}}{\frac{dz}{dx}}\) 

∴ \(\rm \frac{\mathrm{d} y}{\mathrm{d} z}= \frac{\frac{2}{\sqrt{1-x^{2}}}}{\frac{1}{\sqrt{1-x^{2}}}}\) = 2 .

The correct option is 2.