Step 1: Taking natural log of each piece individually:
For \(u=x^{\sin x}\): \(\ln u=\sin x\,\ln x\). For \(v=(\cos x)^{\tan x}\): \(\ln v=\tan x\,\ln(\cos x)\) — treating each term as its own logarithmic-differentiation sub-problem before combining.
Step 2: Implicit differentiation of ln u:
\(\dfrac{u'}{u}=\dfrac{d}{dx}[\sin x\ln x]=\cos x\ln x+\sin x\cdot\dfrac1x\) by the product rule, so \(u'=u\Big(\cos x\ln x+\dfrac{\sin x}{x}\Big)\).
Step 3: Implicit differentiation of ln v:
\(\dfrac{v'}{v}=\dfrac{d}{dx}[\tan x\ln(\cos x)]=\sec^2x\ln(\cos x)+\tan x\cdot\dfrac{-\sin x}{\cos x}\), and \(\tan x\cdot\sin x/\cos x=\tan^2x\), giving \(v'=v[\sec^2x\ln(\cos x)-\tan^2x]\).
Step 4: Adding the two derivatives:
\(y'=u'+v'\), substituting back \(u=x^{\sin x}\) and \(v=(\cos x)^{\tan x}\).
Final Answer:
\[ \boxed{y'=x^{\sin x}\Big(\cos x\ln x+\dfrac{\sin x}{x}\Big)+(\cos x)^{\tan x}\big(\sec^{2}x\ln(\cos x)-\tan^{2}x\big)} \]