Gujarat BoardEnglish MediumSTD 11 ScienceMATHSDerivatives2 Marks
Question
Differentiate the following functions with respect to x:$\frac{\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}$
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Answer
We have, $\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\Big)$
Using Quotient rule, we get
$=\frac{(\sec\text{x}+\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x}\tan\text{x})-(\text{x}\tan\text{x})\frac{\text{d}}{\text{dx}}(\sec\text{x}+\tan\text{x}){}}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-(\text{x}\tan\text{x})(\sec\text{x}\tan\text{x}+\sec^2\text{x})}{(\sec\text{x}+\tan\text{x})^2}$ [Used product rule]
$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-\text{x}\sec\text{x}+\tan^2\text{x}-\text{x}\tan\text{x}\sec^2\text{x})}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-(\text{x}\tan\text{x})(\sec\text{x}\tan\text{x}+\sec^2\text{x})}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-\text{x}\tan\text{x}\sec\text{x}(\sec\text{x}+\tan\text{x})}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\text{x}\sec^2\text{x}+\tan\text{x}-\text{x}\tan\text{x}\sec\text{x})(\sec\text{x}+\tan\text{x})}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\text{x}\sec^2\text{x}+\tan\text{x}-\text{x}\tan\text{x}\sec\text{x})}{(\sec\text{x}+\tan\text{x})}$
$=\frac{\text{x}\sec\text{x}(\sec\text{x}-\tan\text{x})+\tan\text{x}}{(\sec\text{x}+\tan\text{x})}$
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