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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
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Answer

Let $\text{y}=\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3\big]$
$=\frac{\text{d}}{\text{dx}}(\text{x}\sin2\text{x})+\frac{\text{d}}{\text{dx}}(5^\text{x})+\frac{\text{d}}{\text{dx}}(\text{k}^\text{k})+\frac{\text{d}}{\text{dx}}\big(\tan^6\text{x}\big)$
$=\Big[\text{x}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\sin2\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +5^\text{x}\log5+0+6\tan^5\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})$
[Using product rule and chain rule]
$=\Big[\text{x}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\sin2\text{x}\Big]+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
$=2\text{x}\cos2\text{x}+\sin2\text{x}+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{x}\sin2\text{x}+5^\text{x}+\text{k}^\text{k}+(\tan^2\text{x})^3\big) \\ =2\text{x}\cos2\text{x}+\sin2\text{x}+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$

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