Question

Find $\frac{dy}{dx}$

$y={\left(\sin x\right)}^{\cos x}+{\left(\cos x\right)}^{\sin x}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\left(\sin x\right)}^{\cos x}+{\left(\cos x\right)}^{\sin x} \\ \Rightarrow y=e^{\log {\left(\sin x\right)}^{\cos x}}+e^{\log {\left(\cos x\right)}^{\sin x}} \\ \Rightarrow y=e^{\cos x\log \sin x}+e^{\sin x\mathrm{logcos}x} \\ \mathrm{Differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=\frac{d}{dx}\left(e^{\cos x\log \sin x}\right)+\frac{d}{dx}\left(e^{\sin x\mathrm{logcos}x}\right) \\ =e^{\cos x\log \sin x}\frac{d}{dx}\left(\cos x\log \sin x\right)+e^{\sin x\mathrm{logcos}x}\frac{d}{dx}\left(\sin x\mathrm{logcos}x\right) \\ =e^{\log {\left(\sin x\right)}^{\cos x}}\left[\cos x\frac{d}{dx}\log \sin x+\log \sin x\frac{d}{dx}\left(\cos x\right)\right]+e^{\log {\left(\cos x\right)}^{\sin x}}\left[\sin x\frac{d}{dx}\log \cos x+\log \cos x\frac{d}{dx}\left(\sin x\right)\right] \\ ={\left(\sin x\right)}^{\cos x}\left[\cos x\frac{1}{\sin x}\frac{d}{dx}\left(\sin x\right)+\log \sin x\times \left(-\sin x\right)\right]+{\left(\cos x\right)}^{\sin x}\left[\sin x\frac{1}{\cos x}\frac{d}{dx}\left(\cos x\right)+\log \cos x\times \left(\cos x\right)\right] \\ ={\left(\sin x\right)}^{\cos x}\left[\cot x\cos x-\sin x\log \sin x\right]+{\left(\cos x\right)}^{\sin x}\left[\tan x\left(-\sin x\right)+\cos x\log \cos x\right] \\ ={\left(\sin x\right)}^{\cos x}\left[\cot x\cos x-\sin x\log \sin x\right]+{\left(\cos x\right)}^{\sin x}\left[\cos x\log \cos x-\sin x\tan x\right] \end{gathered}$$