Question

Differentiate:

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

Accepted Answer

Given function is $\frac{e^{x} sinx}{secx}$

We know that, from quotient rule ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ , where $v \neq 0$ , $(f g)^{'} = f^{'} g + f g^{'}$ .

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

$u^{'} = \frac{d u}{d x} = \frac{d (e^{x} \sin x)}{d x}$

Applying Product rule $(f g)^{'} = f^{'} g + f g^{'}$ ,

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

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

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

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

Putting the above obtained values in the formula: ${(\frac{u}{v})}^{'} = \frac{u^{'} v - {uv}^{'}}{v^{2}}$

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

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

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

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

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

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

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

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

Therefore, ${[\frac{e^{x} \sin x}{\sec x}]}^{'} = e^{x} (\sin x \cos x + \cos 2 x)$

R S Aggarwal Solutions for Exercise 4: EXERCISE 2D

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 4: EXERCISE 2D

Questions from R S Aggarwal Solutions for Exercise 4: EXERCISE 2D with Hints & Solutions

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