91影视

We have been given

$f\left(x\right)=(\sin 鈦�x{)}^x$

to find the derivative of this function .

Product rule for differentiation

Quotient rule for differentiation

${\left(\frac{f}{g}\right)}^{\mathrm{鈥�}}\left(x\right)=\frac{f^{\mathrm{鈥�}}\left(x\right)g\left(x\right)鈭�f\left(x\right)g^{\mathrm{鈥�}}\left(x\right)}{\left(g\right(x){)}^2}$

Now ,

$\begin{array}{r}\frac{d}{dx}(\sinh 鈦�x)=\cosh 鈦�x\\ \frac{d}{dx}(\cosh 鈦�x)=\sinh 鈦�x\\ \frac{d}{dx}(\tanh 鈦�x)={\mathrm{sech}}^2鈦�x\\ \frac{d}{dx}\left({\sinh }^{鈭�1}鈦�x\right)=\frac{1}{\sqrt{x^2+1}}\\ \frac{d}{dx}\left({\cosh }^{鈭�1}鈦�x\right)=\frac{1}{\sqrt{x^2鈭�1}}\\ \frac{d}{dx}\left({\tanh }^{鈭�1}鈦�x\right)=\frac{1}{1鈭�x^2}\end{array}$

$\begin{array}{r}f\left(x\right)=(\sin 鈦�x{)}^x\\ \ln 鈦�f\left(x\right)=\ln 鈦�(\sin 鈦�x{)}^x\text{[Taking log in both sides]}\\ \ln 鈦�f\left(x\right)=x\ln 鈦�(\sin 鈦�x)\\ \frac{f^{\mathrm{鈥�}}\left(x\right)}{f\left(x\right)}=\frac{d}{dx}(x\ln 鈦�(\sin 鈦�x\left)\right)\left[\begin{array}{c}\text{Differentiate both sides}\\ \text{with respect to}x\end{array}\right]\\ \frac{f^{\mathrm{鈥�}}\left(x\right)}{f\left(x\right)}=\left(\frac{d}{dx}\left(x\right)\right)\ln 鈦�(\sin 鈦�x)+x\left(\frac{d}{dx}\ln 鈦�(\sin 鈦�x)\right)\\ =\ln 鈦�(\sin 鈦�x)+\frac{x}{\sin 鈦�x}\left(\frac{d}{dx}\sin 鈦�x\right)\left[\begin{array}{c}\text{Using product rule}\\ \text{Using chain rule}\\ \frac{d}{dx}\ln 鈦�x=\frac{1}{x}\end{array}\right]\\ =\ln 鈦�(\sin 鈦�x)+\frac{x}{\sin 鈦�x}(\cos 鈦�x)\left[\frac{d}{dx}\sin 鈦�x=\cos 鈦�x\right]\\ f^{\mathrm{鈥�}}\left(x\right)=\left(\ln 鈦�(\sin 鈦�x)+\frac{x\cos 鈦�x}{\sin 鈦�x}\right)f\left(x\right)\left[\text{Multiplying both sides by}f\right(x\left)\right]\\ f^{\mathrm{鈥�}}\left(x\right)=\left(\ln 鈦�(\sin 鈦�x)+\frac{x\cos 鈦�x}{\sin 鈦�x}\right)(\sin 鈦�x{)}^x\end{array}$