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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:
$(\sin\text{x})^{\cos\text{x}}$
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Answer

Let $\text{y}=(\sin\text{x})^{\cos\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=(\sin\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{y}=\cos\text{x}\log(\sin\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\sin\text{x}}(\cos\text{x})-\sin\text{x}\log\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}[\cos\text{x}\cot\text{x}-\sin\text{x}\log\sin\text{x}]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}[\cos\text{x}\cot\text{x}-\sin\text{x}\log\sin\text{x}]$

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