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