Continuity and Differentiability — Class 12 Maths Solution

Let $y = \cos {x^3} \cdot {\sin ^2}({x^5})$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}[\cos {x^3} \cdot {\sin ^2}({x^5})]$

$= \cos {x^3}\cfrac{d}{{dx}}{\sin ^2}({x^5}) + {\sin ^2}({x^5})\cfrac{d}{{dx}}\cos {x^3}$

$= \cos {x^3} \cdot 2\sin ({x^5})\cfrac{d}{{dx}}\sin ({x^5}) + {\sin ^2}({x^5})( - \sin {x^3})\cfrac{d}{{dx}}({x^3})$

$= \cos {x^3} \cdot 2\sin ({x^5})\cos ({x^5})\cfrac{d}{{dx}}({x^5}) + {\sin ^2}({x^5})( - \sin {x^3})(3{x^2})$

$= 10{x^4}\cos {x^3}\sin ({x^5})\cos ({x^5}) - 3{x^2}{\sin ^2}({x^5})\sin {x^3}$ .

therefore $\frac{dy}{dx}=10{x^4}\cos {x^3}\sin ({x^5})\cos ({x^5}) - 3{x^2}{\sin ^2}({x^5})\sin {x^3}$