# Question #30f2b

Mar 7, 2017

$\left(1\right) : \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left\{5 \cos 5 x - \cos x\right\} , \mathmr{and} , \left(1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cos 3 x \cos 2 x - 3 \sin 3 x \sin 2 x .$

$\left(2\right) : \frac{\mathrm{dY}}{\mathrm{dx}} = 6 \left(x - 1\right) {\left(x - 3\right)}^{2} {\left(x + 1\right)}^{2.}$

#### Explanation:

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

Let, $\left({x}^{2} - 2 x - 3\right) = t , s o , Y = {t}^{3} , t = {x}^{2} - 2 x - 3.$

Thus, $Y$ is a fun. of $t ,$ &, $t$ of $x .$

In such cases, $\frac{\mathrm{dY}}{\mathrm{dx}}$ can be obtained by The Chain Rule :-

$\text{The Chain Rule : } \frac{\mathrm{dY}}{\mathrm{dx}} = \left(\frac{\mathrm{dY}}{\mathrm{dt}}\right) \left(\frac{\mathrm{dt}}{\mathrm{dx}}\right) \ldots \ldots \ldots \ldots \left(\star\right) .$

Recall that, $\frac{d}{\mathrm{dt}} \left({t}^{n}\right) = n {t}^{n - 1} .$

Hence, $Y = {t}^{3} \Rightarrow \frac{\mathrm{dY}}{\mathrm{dt}} = 3 {t}^{2.} . . \left(1\right) , \mathmr{and} , t = {x}^{2} - 2 x - 3$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}} = 2 x - 2 = 2 \left(x - 1\right) \ldots \ldots \ldots \ldots . \left(2\right) .$

Using $\left(1\right) \mathmr{and} \left(2\right) \in \left(\star\right) ,$ we get,

$\frac{\mathrm{dY}}{\mathrm{dx}} = \left(3 {t}^{2}\right) \left\{2 \left(x - 1\right)\right\} = 6 \left(x - 1\right) {t}^{2} ,$ &, returning back from $t$ to $x$,

$\frac{\mathrm{dY}}{\mathrm{dx}} = 6 \left(x - 1\right) {\left({x}^{2} - 2 x - 3\right)}^{2} = 6 \left(x - 1\right) {\left\{\left(x - 3\right) \left(x + 1\right)\right\}}^{2} , \mathmr{and} ,$

$\frac{\mathrm{dY}}{\mathrm{dx}} = 6 \left(x - 1\right) {\left(x - 3\right)}^{2} {\left(x + 1\right)}^{2.}$

Regarding, the Diffn. of $y = \sin 2 x \cos 3 x ,$ we have, $2$ Methods :-

Method 1:-

We know, that, $2 \cos A \sin B = \sin \left(A + B\right) - \sin \left(A - B\right) .$

With, $A = 3 x , B = 2 x , 2 \cos 3 x \sin 2 x = \sin \left(3 x + 2 x\right) - \sin \left(3 x - 2 x\right)$

$\therefore y = \cos 3 x \sin 2 x = \frac{1}{2} \left\{\sin 5 x - \sin x\right\} .$

Therefore, using the Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left\{\left(\cos 5 x\right) \frac{d}{\mathrm{dx}} \left(5 x\right) - \cos x\right\} , \mathmr{and} ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left\{5 \cos 5 x - \cos x\right\} .$

Method 2:-

In this Method, we use the following Product Rule, together

with the Chain Rule.

$\text{Product Rule : } \frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}} .$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\cos 3 x\right) \frac{d}{\mathrm{dx}} \left(\sin 2 x\right) + \left(\sin 2 x\right) \frac{d}{\mathrm{dx}} \left(\cos 3 x\right)$

$= \left(\cos 3 x\right) \left\{\left(\cos 2 x\right) \frac{d}{\mathrm{dx}} \left(2 x\right)\right\} + \left(\sin 2 x\right) \left\{\left(- \sin 3 x\right) \frac{d}{\mathrm{dx}} \left(3 x\right)\right\}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cos 3 x \cos 2 x - 3 \sin 3 x \sin 2 x .$

I leave it upon the reader to show that both the Answers are same.

Enjoy Maths.!