If y=e^ x^2 +( )^ x , then find dy dx . - Vidyadip

Question

If $\text{y}=\text{e}^{\text{x}^2\cos\text{x}}+(\cos\text{x})^{\text{x}}, $ then find $\frac{\text{dy}}{\text{dx}}.$

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Answer

Let $\text{u}=(\cos\text{x})^{\text{x}}\Rightarrow\text{y}=\text{e}^{\text{x}^2\cos\text{x}}+\text{u}$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^2\cos\text{x}}(2\text{x}\cdot\cos\text{x}-\text{x}^2\cdot\sin\text{x})+\frac{\text{du}}{\text{dx}}$
$\log\text{u}=\log(\cos\text{x})^\text{x}\Rightarrow\log\text{u}=\text{x}\cdot\log(\cos\text{x})$
Differentiate w.r.t. “x”
$\frac{\text{1}}{\text{u}}\frac{\text{du}}{\text{dx}}=\log(\cos\text{x})-\text{x}\tan\text{x}$ $\Rightarrow\frac{\text{du}}{\text{dx}}=(\cos\text{x})^{\text{x}}\{\log(\cos\text{x})-\text{x}\tan\text{x}\}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^2\cos\text{x}}(2\text{x}\cdot\cos\text{x}-\text{x}^2\cdot\sin\text{x})\\+(\cos\text{x})^{\text{x}}\{\log(\cos\text{x})-\text{x}\tan\text{x}\}$

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