# Differentiate

Question:

Differentiate

$\frac{e^{x} \sin x}{\sec x}$

Solution:

To find: Differentiation of $\left(\frac{e^{x} \sin x}{\sec x}\right)$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d \sin x}{d x}=\cos x$

(iii) $\frac{d \sec x}{d x}=\sec x \tan x$

(iv) $\frac{d e^{x}}{d x}=e^{x}$

(v) $(\text { uv })^{\prime}=u^{\prime} v+u v^{\prime}($ Leibnitz or product rule)

Let us take $u=\left(e^{x} \sin x\right)$ and $v=(\sec x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(e^{x} \sin x\right)}{d x}$

Applying Product rule

$(g h)^{\prime}=g^{\prime} h+g h^{\prime}$

Taking $g=e^{x}$ and $h=\sin x$

$=e^{x} \sin x+e^{x} \cos x$

$u^{\prime}=e^{x} \sin x+e^{x} \cos x$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\sec x)}{d x}=\sec x \tan x$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

$\left[\frac{e^{x} \sin x}{\sec x}\right]^{\prime}=\frac{\left(e^{x} \sin x+e^{x} \cos x\right) \times(\sec x)-\left(e^{x} \sin x\right) \times(\sec x \tan x)}{(\sec x)^{2}}$

$=\frac{\left(e^{x} \sin x+e^{x} \cos x\right)-\left(e^{x} \sin x\right) \times(\tan x)}{(\sec x)}$

$=\cos x\left[\left(e^{x} \sin x+e^{x} \cos x\right)-\left(e^{x} \sin x\right) x(\tan x)\right]$

$=\left[\left(e^{x} \sin x \cos x+e^{x} \cos ^{2} x\right)-\left(e^{x} \sin x \cos x\right) x(\tan x)\right]$

$=\left[\left(e^{x} \sin x \cos x+e^{x} \cos ^{2} x\right)-\left(e^{x} \sin ^{2} x\right)\right]$

$=\left(e^{x} \sin x \cos x+e^{x} \cos ^{2} x-e^{x} \sin ^{2} x\right.$

$=\left(e^{x} \sin x \cos x+e^{x} \cos ^{2} x-e^{x} \sin ^{2} x\right.$

$=e^{x} \sin x \cos x+e^{x} \cos 2 x$

$=e^{x}(\sin x \cos x+\cos 2 x)$

Ans $)=e^{x}(\sin x \cos x+\cos 2 x)$