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UPSC NDA Math Model Paper 6

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Question : 37 of 120

Marks: +1, -0

Derivative of

\tan^3 \theta

\sec^2 \theta = \frac{\pi}{3}

is

$\frac{3}{2}$
$\frac{3 \sqrt{3}}{2}$
$\frac{1}{2}$
$-\frac{\sqrt{3}}{2}$

Solution:

Consider,

y = \tan^3 \theta = f(\theta)

x = \sec^2 \theta = f(\theta)

To find

\frac{dy}{dx}

, use chain rule

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

........(1)

Here,

y = \tan^3 \theta

\frac{dy}{d\theta} = 3 \tan^2 \theta \sec^2 \theta

Now,

x = \sec^2 \theta

\frac{dx}{d\theta} = 2 \sec^2 \theta \tan \theta

Put these values in equation (1), we get

\frac{dy}{dx} = \frac{3}{2} \tan \theta

To find

\frac{dy}{dx}

at

\theta = \frac{\pi}{3}

Put

\theta = \frac{\pi}{3}

in above equation.

\frac{dy}{dx} = \frac{3 \sqrt{3}}{2}

Hence, Derivative of

\tan^3 \theta

w.r.t. to

\sec^2 \theta = \frac{\pi}{3}

is

\frac{dy}{dx} = \frac{3 \sqrt{3}}{2}

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