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Question
Differentiate \[x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3\] ?
Solution
\[\text{Let} y = x \sin2x + 5^x + k^k + \left( \tan^2 x \right)^3\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = \frac{d}{dx}\left[ x \sin2x + 5^x + k^k + \left( \tan^6 x \right) \right]\]
\[ = \frac{d}{dx}\left( x \sin2x \right) + \frac{d}{dx}\left( 5^x \right) + \frac{d}{dx}\left( k^k \right) + \frac{d}{dx}\left( \tan^6 x \right)\]
\[ = \left[ x\frac{d}{dx}\left( \sin2x \right) + \sin2x\frac{d}{dx}\left( x \right) \right] + 5^x \log_e 5 + 0 + 6 \tan^5 x \times \frac{d}{dx}\left( \tan x \right) \left[ \text{Using product rule and chain rule } \right]\]
\[ = \left[ x \cos2x\frac{d}{dx}\left( 2x \right) + \sin2x \right] + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x\]
\[ = 2x \cos2x + \sin2x + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x\]
\[So, \frac{d}{dx}\left\{ x \sin2x + 5^x + k^k + \left( \tan^2 x \right)^3 \right\} = 2x\cos2x + \sin2x + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x\]
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