Chapter 13: Limits and Derivatives

Ans: Given:

\[{\sin ^3}x{\cos ^3}x\]

We know that, Differentiate the given equation w.r.t \[x\] using product rule and chain rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

$\because \sin 2x = 2\sin x\cos x$

Let \[y = {\sin ^3}x{\cos ^3}x\]

\[\dfrac{{dy}}{{dx}} = {\sin ^3}x \cdot \dfrac{d}{{dx}}{\cos ^3}x + {\cos ^3}x\dfrac{d}{{dx}}{\sin ^3}x\]                    (by product rule)

$= {\sin ^3}x \cdot 3{\cos ^2}x( - \sin x) + {\cos ^3}x \cdot 3{\sin ^2}x\cos x\quad$ (by chain rule)

$=  - 3{\cos ^2}x{\sin ^4}x + 3{\sin ^2}x{\cos ^4}x$

$= 3{\sin ^2}x{\cos ^2}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)$

$= 3{\sin ^2}x{\cos ^2}x\cos 2x$

$= \dfrac{3}{4}{(2\sin x\cos x)^2}\cos 2x{\text{                                       }}\left[ {\because \sin 2x = 2\sin x\cos x} \right]$

$= \dfrac{3}{4}{\sin ^2}2x\cos 2x$

Ans: Given:

$\dfrac{1}{{a{x^2} + bx + c}}$

We know that, Differentiate the given equation w.r.t \[x\] using product rule and chain rule

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

Let $y = \dfrac{1}{{a{x^2} + bx + c}} = {\left( {a{x^2} + bx + c} \right)^{ - 1}}$

$\dfrac{{dy}}{{dx}} =  - {\left( {a{x^2} + bx + c} \right)^{ - 2}}\left( {2ax + b} \right){\text{ }}\left[ {{\text{By chain rule}}} \right]$

$= \dfrac{{ - (2ax + b)}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}}$