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Answer: $2 x \cos 2 x+\sin 2 x+5^{x} \log _{e} 5+6 \tan ^{5} x \sec ^{2} x$

Hint: You must know about the rules of solving derivative of trigonometric function.

Given:  $x \sin 2 x+5^{x}+k^{k}+\left(\tan ^{6} x\right)$

Solution:

Let  $y=x \sin 2 x+5^{x}+k^{k}+\left(\tan ^{6} x\right)$

Differentiate with respect to x,

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{x} \sin 2 \mathrm{x}+5^{\mathrm{x}}+\mathrm{k}^{\mathrm{k}}+\left(\tan ^{6} \mathrm{x}\right)\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xsin} 2 \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}\left(5^{\mathrm{x}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{k}^{\mathrm{k}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{6} \mathrm{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\mathrm{x}\left\{\cos 2 \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x})\right\}+\sin 2 \mathrm{x}\right]+5^{\mathrm{x}} \log _{\mathrm{e}} 5+6 \tan ^{5} \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}}(\tan \mathrm{x})$

$\frac{d y}{d x}=\left[x\left\{\cos 2 x \frac{d}{d x}(2 x)\right\}+\sin 2 x\right]+5^{x} \log _{e} 5+6 \tan ^{5} x \sec ^{2} x$

$\frac{d y}{d x}=2 x \cos 2 x+\sin 2 x+5^{x} \log _{e} 5+6 \tan ^{5} x \sec ^{2} x$

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