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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x:
$\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x}$
✓

Answer

Let $\text{y}=\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x})$
$=\text{e}^{\text{ax}}\frac{\text{d}}{\text{dx}}\big\{\sec\text{x}\tan2\text{x}\big\}+\sec\text{x}\tan2\text{x}\frac{\text{d}}{\text{dx}}\big\{\text{e}^{\text{ax}}\big\}$
$=\text{e}^{\text{ax}}\big[\text{sec}\text{x}\tan\text{x}\tan2\text{x}+2\sec^2 2\text{x}\sec\text{x}\big]+\text{ae}^{\text{ax}}\sec\text{a}\tan^{2\text{x}}$
$=\text{ae}^{\text{ax}}\sec\text{x}\tan2\text{x}+\text{e}^{\text{ax }}\sec\text{x}\tan\text{x}\tan2\text{x}+2\text{e}^{\text{ax}}\sec\text{x}\sec^2 2\text{x}$
$=\text{e}^{\text{ax}}\sec\text{x}\big\{\text{a}\tan2\text{x}+\tan\text{x}\tan2\text{x}+2\sec^22\text{x}\big\}$
So,
$\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x})=\text{e}^{\text{ax}}\sec\text{x}\big\{\text{a}\tan2\text{x}+\tan\text{x}\tan2\text{x}+2\sec^22\text{x}\big\}$

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