Given that $\displaystyle \mathrm{y}=(\sin x)^{x} \cdot x^{\sin x}+\mathrm{a}^{x}$, find $\displaystyle \frac{\mathrm{dy}}{\mathrm{d} x}$.

As, $\displaystyle \mathrm{y}=(\sin x)^{x} \cdot x^{\sin x}+\mathrm{a}^{x}=u+\mathrm{a}^{x} \Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\frac{\mathrm{d} u}{\mathrm{d} x}+\frac{\mathrm{d}\left(\mathrm{a}^{x}\right)}{\mathrm{d} x}$ where $\displaystyle u=(\sin x)^{x} . x^{\sin x} \Rightarrow \log u=x \log (\sin x)+\sin x . \log x$ on differentiating both sides with respect to $\displaystyle x$, we get\Rightarrow \frac{\mathrm{d} u}{\mathrm{d} x}=(\sin x)^{x} \cdot x^{\sin x}\left[\log (\sin x)+x \cot x+\frac{\sin x}{x}+\log x \cdot \cos x\right]Thus, $\displaystyle \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=(\sin x)^{x} \cdot x^{\sin x}\left[\log (\sin x)+x \cot x+\frac{\sin x}{x}+\log x \cdot \cos x\right]+\mathrm{a}^{x} \log \mathrm{a}$