Question

Evaluate:

$\frac{d}{dx}[{\left(cosx\right)}^{logx}+{\left(logX\right)}^{{}^x}]$=

Answer 1

We have,

$f\left(x\right)=\frac{d}{dx}[{\left(\cos x\right)}^{\log x}+{\left(\log x\right)}^x]$

Let,

$y_1={\left(\cos x\right)}^{\log x}......\left(1\right)$

$\text{and}$

$y_2={\left(\log x\right)}^x.......\left(2\right)$

$f\left(x\right)=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}.......\left(A\right)$

Taking log both side and we get,

$\log y_1=\log {\left(\cos x\right)}^{\log x}$

$\log y_1=\log x.\log \left(\cos x\right)......\left(1\right)$

$\log y_2=\log {\left(\log x\right)}^x$

$\log y_2=x\log \left(\log x\right).......\left(2\right)$

Differentiation equation (1) with respect to x and we get,

$\frac{d\log y_1}{dx}=\frac{d}{dx}(\log x.\log \cos x)$

$\frac{1}{y_1}\frac{d{y}_1}{dx}=\log x\frac{d}{dx}\log \cos x+\log \cos x\frac{d}{dx}\log x$

$\frac{1}{y_1}\frac{d{y}_1}{dx}=\log x\frac{1}{\cos x}\sin x+\frac{\log \cos x}{x}$

$\frac{1}{y_1}\frac{d{y}_1}{dx}=\tan x\log x+\frac{\log \cos x}{x}$

$\frac{d{y}_1}{dx}=y_1(\tan x\log x+\frac{\log \cos x}{x})$

$\frac{d{y}_1}{dx}={\left(\cos x\right)}^{\log x}(\tan x\log x+\frac{\log \cos x}{x})$

Differentiation equation (2) with respect to x and we get,

$\frac{d}{dx}\log y_2=\frac{d}{dx}x.\log \log x$

$\frac{1}{y_2}\frac{d{y}_2}{dx}=x\frac{1}{\log x}\frac{1}{x}+\log \log x\left(1\right)$

$\frac{1}{y_2}\frac{d{y}_2}{dx}=\frac{1}{\log x}+\log \log x$

$\frac{d{y}_2}{dx}=y_2(\frac{1}{\log x}+\log \log x)$

$\frac{d{y}_2}{dx}={\left(\log x\right)}^x(\frac{1}{\log x}+\log \log x)$

Put the value of $\frac{d{y}_1}{dx}and\frac{d{y}_2}{dx}$ in equation (A) and we get,

$f\left(x\right)={\left(\cos x\right)}^{\log x}(\tan x\log x+\frac{\log \cos x}{x})+{\left(\log x\right)}^x(\frac{1}{\log x}+\log \log x)$Hence, this is the answer,