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Differentiate Tan (X° + 45°) ?

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Question

Differentiate tan (x° + 45°) ?

Sum
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Solution

\[Let, y = \tan\left( x^\circ+ 45^\circ \right)\]
\[ \Rightarrow y = \tan\left\{ \left( x + 45 \right)\frac{\pi}{180} \right\}\]
\[\text{ Differentiating it with respect to x we get }, \]
\[\frac{d y}{d x} = \frac{d}{dx}\tan\left\{ \left( x + 45 \right)\frac{\pi}{180} \right\}\]
\[ = \sec^2 \left\{ \left( x + 45 \right)\frac{\pi}{180} \right\} \times \frac{d}{dx}\left( x + 45 \right)\frac{\pi}{180}\] ...........[Using chain rule]
\[ = \frac{\pi}{180} \sec^2 \left( x^\circ + 45^\circ \right)\]
So,
\[ \frac{d}{dx}\left\{ \tan\left( x^\circ+ 45^\circ \right) \right\} = \frac{\pi}{180} \sec^2 \left( x^\circ + 45^\circ \right)\]

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Chapter 10: Differentiation - Exercise 11.02 [Page 37]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 10 Differentiation
Exercise 11.02 | Q 3 | Page 37
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