Question
Differentiate the following functions with respect to x:
$\sin(\text{x}^\text{x})$
$\sin(\text{x}^\text{x})$
✓
Answer
Let $\text{y}=\sin(\text{x}^\text{x})\ .....(\text{i})$
Taking log on both sides,
$\log(\sin^{-1}\text{y})=\log\text{x}^\text{x}$
$\Rightarrow\ \log(\sin^{-1}\text{y})=\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\frac{\text{dy}}{\text{dy}}(\sin^{-1}\text{y})=\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\times\Big(\frac{1}{\sqrt{1-\text{y}^2}}\Big)\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{y}\sqrt{1-\text{y}^2}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}(\sin\text{x}^\text{x})\sqrt{1-(\sin\text{x}^\text{x})^2}(1+\log\text{x})$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\cos\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
Taking log on both sides,
$\log(\sin^{-1}\text{y})=\log\text{x}^\text{x}$
$\Rightarrow\ \log(\sin^{-1}\text{y})=\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\frac{\text{dy}}{\text{dy}}(\sin^{-1}\text{y})=\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\times\Big(\frac{1}{\sqrt{1-\text{y}^2}}\Big)\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{y}\sqrt{1-\text{y}^2}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}(\sin\text{x}^\text{x})\sqrt{1-(\sin\text{x}^\text{x})^2}(1+\log\text{x})$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\cos\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
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