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Differentiate E a X Sec X Tan 2 X ? - Mathematics

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प्रश्न

Differentiate \[e^{ax} \sec x \tan 2x\] ?

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उत्तर

\[\text{Let } y = e^{ax} \sec x \tan2x\]

Differentiate it with respect to x,

\[\frac{d y}{d x} = \frac{d}{dx}\left( e^{ax} sec x \tan2x \right)\]

\[ = e^{ax} \frac{d}{dx}\left\{ \sec x \tan2x \right\} + \sec x \tan2x\frac{d}{dx}\left\{ e^{ax} \right\} \]

\[ = e^{ax} \left[ \sec x \tan x \tan2x + 2 \sec^2 2x\sec x \right] + a e^{ax} \sec x \tan2x \]

\[ = e^{ax} \left[ \sec x \tan x \tan2x + 2 \sec^2 2x\sec x \right] + a e^{ax} sec x \tan2x\]

\[ = a e^{ax} \sec x \tan2x + e^{ax} \sec x \tan x \tan2x + 2 e^{ax} \sec x \sec^2 2x\]

\[ = e^{ax} \sec x\left\{ a \tan2x + \tan x \tan2x + 2 \sec^2 2x \right\}\]

\[So, \frac{d}{dx}\left( e^{ax} \sec x \tan2x \right) = e^{ax} \sec x\left\{ a \tan2x + \tan x \tan2x + 2 \sec^2 2x \right\}\]

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अध्याय 11: Differentiation - Exercise 11.02 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.02 | Q 54 | पृष्ठ ३८

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