# Find the values

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

$1 y=e^{x}+10^{x}+x^{x}$

Solution:

let $y=e^{x}+10^{x}+x^{x}$

$\Rightarrow y=a+b+c$

where $\mathrm{a}=\mathrm{e}^{\mathrm{x}} ; \mathrm{b}=10^{\mathrm{x}} ; \mathrm{c}=\mathrm{x}^{\mathrm{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}+\frac{\mathrm{dc}}{\mathrm{dx}}$

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where a and $\mathrm{u}$ are any variables $\}$

$a=e^{x}$

Taking log both the sides:

$\Rightarrow \log a=\log e^{x}$

$\Rightarrow \log a=x \log e$

$\left\{\log x^{a}=\operatorname{alog} x\right\}$

$\Rightarrow \log a=x\{\log e=1\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{dx}}{\mathrm{dx}}$

$\Rightarrow \frac{1}{\mathrm{a}} \frac{\mathrm{da}}{\mathrm{dx}}=1$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=\mathrm{a}$

Put the value of $a=e^{x}$

$\Rightarrow \frac{\mathrm{da}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$

$\mathrm{b}=10^{\mathrm{x}}$

Taking log both the sides:

$\Rightarrow \log \mathrm{b}=\log 10^{x}$

$\Rightarrow \log \mathrm{b}=x \log 10$

$\left\{\log x^{a}=a \log x\right\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{x} \log 10)}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{b})}{\mathrm{dx}}=\log 10 \times \frac{\mathrm{dx}}{\mathrm{dx}}$

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{au})}{\mathrm{dx}}=\mathrm{a} \frac{\mathrm{du}}{\mathrm{dx}}$ where a is any constant and $\mathrm{u}$ is any variable $\}$

$\Rightarrow \frac{1}{b} \frac{d b}{d x}=b(\log 10)$

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x}\right\}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\mathrm{b}(\log 10)$

Put the value of $b=10^{x}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=10^{\mathrm{x}}(\log 10)$

$c=x^{x}$

Taking log both the sides:

$\Rightarrow \log c=\log x^{x}$

$\Rightarrow \log c=x \log x$

$\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{c})}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{x} \log \mathrm{x})}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{c})}{\mathrm{dx}}=\mathrm{x} \times \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}+\log \mathrm{x} \times \frac{\mathrm{dx}}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{c} \frac{d c}{d x}=x \times \frac{1}{x} \frac{d x}{d x}+\log x$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{c} \frac{d c}{d x}=1+\log x$

$\Rightarrow \frac{d c}{d x}=c\{1+\log x\}$

Put the value of $c=x^{x}$

$\Rightarrow \frac{\mathrm{dc}}{\mathrm{dx}}=\mathrm{x}^{\mathrm{x}}\{1+\log \mathrm{x}\}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}+\frac{\mathrm{dc}}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}+10^{\mathrm{x}}(\log 10)+\mathrm{x}^{\mathrm{x}}\{1+\log \mathrm{x}\}$