Differentiate the functions given in Exercise: x^x-2^ - Vidyadip

Question

Differentiate the functions given in Exercise:
$\text{x}^\text{x}-2^{\sin\text{x}}$

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Answer

Let $\text{y}=\text{x}^\text{x}-2^{\sin\text{x}}$
Putting $\text{u}=\text{x}^\text{x}\text{ and v }=2^{\sin\text{x}}$
$\Rightarrow\ \text{y}=\text{u}-\text{v}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}-\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now, $u = x^x$ $\ \Rightarrow\ \log\text{u}=\log\text{x}^\text{x}=\text{x}\log\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\text{x}}+\log\text{x}.1$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}(1+\log\text{x})$ $=\text{x}^\text{x}(1+\log\text{x})\ \dots\text{(ii)}$
Again, $\text{v}=2^{\sin\text{x}}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}2^{\sin\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=2^{\sin \text{x}}\log2\frac{\text{d}}{\text{dx}}\sin\text{x}=\ \Big[\because\frac{\text{d}}{\text{dx}}\text{a}^{\text{f(x)}}=\text{a}^{\text{f(x)}}\log\text{a}\frac{\text{d}}{\text{dx}}\text{f(x)}\Big]$
$\frac{\text{dv}}{\text{dx}}=2^{\sin\text{x}}(\log2).\cos\text{x}=\cos\text{x}.2^{\sin\text{x}}\log2\ \dots\text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})-\cos\text{x}.2^{\sin\text{x}}\log2$

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