Continuity and Differentiability — Class 12 Maths Solution

Let $y = {\sin ^n}\left( {a{x^2} + bx + c} \right)$

therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left[ {\sin \left( {a{x^2} + bx + c} \right)} \right]^n}$

$= n \cdot {\left[ {\sin \left( {a{x^2} + bx + c} \right)} \right]^{n - 1}} \cdot \frac{d}{{dx}}\sin \left( {a{x^2} + bx + c} \right)$

$= n \cdot {\sin ^{n - 1}}\left( {a{x^2} + bx + c} \right) \cdot \cos \left( {a{x^2} + bx + c} \right) \cdot \frac{d}{{dx}}\left( {a{x^2} + bx + c} \right)$

$= n \cdot {\sin ^{n - 1}}\left( {a{x^2} + bx + c} \right) \cdot \cos \left( {a{x^2} + bx + c} \right) \cdot (2ax + b)$

$= n \cdot (2ax + b) \cdot {\sin ^{n - 1}}\left( {a{x^2} + bx + c} \right) \cdot \cos \left( {a{x^2} + bx + c} \right)$

$\Rightarrow \frac{dy}{dx} = n \cdot (2ax + b) \cdot {\sin ^{n - 1}}\left( {a{x^2} + bx + c} \right) \cdot \cos \left( {a{x^2} + bx + c} \right)$