Question
Differentiate the following functions with respect to x:
$(1+\cos\text{x})^\text{x}$
$(1+\cos\text{x})^\text{x}$
✓
Answer
Let $\text{y}=(1+\cos\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(1+\cos\text{x})^\text{x}$
$\log\text{y}=\text{x}\log(1-\cos\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(1+\cos\text{x})+\log(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{(1+\cos\text{x})}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})+\log(1+\cos\text{x})(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{(1+\cos\text{x})}(0-\sin\text{x})+\log(1+\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\cos\text{x})^\text{x}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
[Using equation (i)]
Taking log on both the sides,
$\log\text{y}=\log(1+\cos\text{x})^\text{x}$
$\log\text{y}=\text{x}\log(1-\cos\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(1+\cos\text{x})+\log(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{(1+\cos\text{x})}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})+\log(1+\cos\text{x})(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{(1+\cos\text{x})}(0-\sin\text{x})+\log(1+\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\cos\text{x})^\text{x}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
[Using equation (i)]
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