Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\sin^2\{\log(2\text{x}+3)\}$
✓
Answer
Consider $\text{y}=\sin^2\{\log(2\text{x}+3)\}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^2\{\log(2\text{x}+3)\}\big]$
$=2\sin\big(\log(2\text{x}+3)\big)\frac{\text{d}}{\text{dx}}\sin\big(\log(2\text{x}+3)\big)$
[Using chain rule]
$=2\sin\big(\log(2\text{x}+3)\big)\cos\big(\log(2\text{x}+3)\big)\frac{\text{d}}{\text{dx}}(2\text{x}+3)$
$=\sin\big(2\log(2\text{x}+3)\big)\times\frac{1}{(2\text{x}+3)}\frac{\text{d}}{\text{dx}}(2\text{x}+3)$
$\big[\text{Since}, 2\sin\text{A}\cos\text{A}=\sin^2\text{A}\big]$
$=\sin\big(2\log(2\text{x}+3)\big)\times\frac{2}{(2\text{x}+3)}$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(\sin^2\log(2\text{x}+3)\big)=\sin\big(2\log(2\text{x}+3)\big)\times\frac{2}{(2\text{x}+3)}$
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