Question Number 162280 by MathsFan last updated on 28/Dec/21
$${y}={x}^{{sinx}} \\ $$$${find}\:\:{y}' \\ $$
Answered by Ar Brandon last updated on 28/Dec/21
$${y}={x}^{\mathrm{sin}{x}} \\ $$$$\mathrm{ln}{y}=\mathrm{sin}{x}\mathrm{ln}{x} \\ $$$$\frac{\mathrm{1}}{{y}}\centerdot\frac{{dy}}{{dx}}=\frac{\mathrm{sin}{x}}{{x}}+\mathrm{cos}{x}\mathrm{ln}{x} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={x}^{\mathrm{sin}{x}} \left(\frac{\mathrm{sin}{x}}{{x}}+\mathrm{cos}{x}\mathrm{ln}{x}\right) \\ $$