CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Differentiate the following functions with respect to x: $\tan5\text{x}^\circ$
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Answer
Let, $\text{y}=\tan5\text{x}^\circ$ $\Rightarrow\ \text{y}=\tan\Big(5\text{x}\times\frac{\pi}{180}\Big)$ Differentiate it with respect to x we get, $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan\Big(5\text{x}\times\frac{\pi}{180}\Big)$ $=\sec^2\Big(5\text{x}\times\frac{\pi}{180}\Big)\frac{\text{d}}{\text{dx}}\Big(5\text{x}\times\frac{\pi}{108}\Big) $ [Using chain rule] $=\Big(\frac{5\text{x}}{180}\Big)\sec^2\Big(5\text{x}\times\frac{\pi}{180}\Big)$ $=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$ Hence, $\frac{\text{d}}{\text{dx}}(\tan5\text{x}^\circ)=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$
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