Find dy dx y = ex + 10x + xx - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$
y = e^x + 10^x + x^x

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Answer

Here, $\text{y}=\text{e}^\text{x}+10^\text{x}+\text{x}^\text{x}$
$\text{e}^{\text{x}}+10^{\text{x}}+\text{e}^{\log\text{x}^\text{x}}$
$\big[\text{Since, e}^{{\log}_\text{a}^\text{b}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\text{y}=\text{e}^{\text{x}}+10^{\text{x}}+\text{e}^{\log\text{x}^\text{x}}$
Differentiating it with respect to x using product rule, chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\frac{\text{d}}{\text{dx}}(10^{\text{x}})+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\times\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^{\text{x}}[1+\log\text{x}]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^{\text{x}}[\log\text{e}+\log\text{x}] \big[\text{Since}, \log_\text{e}\text{e}=1\big]$
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^\text{x}(\log\text{ex})\ \big[\text{Since}\log\text{A}+\log\text{B}=\log\text{AB}]$

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