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Differentiate Tan 5x° ? - Mathematics

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प्रश्न

Differentiate tan 5x° ?

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उत्तर

\[\text{ Let } y = \tan5x^\circ\]

\[ \Rightarrow y = \tan\left( 5x \times \frac{\pi}{180} \right)\]

\[\text{ Differentiate it with respect to x we get }, \]

\[\frac{d y}{d x} = \frac{d}{dx}\tan\left( 5x \times \frac{\pi}{180} \right)\]

\[ = \sec^2 \left( 5x \times \frac{\pi}{180} \right)\frac{d}{dx}\left( 5x \times \frac{\pi}{180} \right) \left[ \text{ using chain rule } \right]\]

\[ = \left( \frac{5\pi}{180} \right) \sec^2 \left( 5x \times \frac{\pi}{180} \right)\]

\[ = \frac{5\pi}{180} \sec^2 \left( 5x^\circ\right)\]

\[\text{Hence}, \frac{d}{dx}\left( \tan5x^\circ \right) = \frac{5\pi}{180} \sec^2 \left( 5x^\circ \right)\]

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पाठ 11: Differentiation - Exercise 11.02 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.02 | Q 9 | पृष्ठ ३७

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