1. **Problem statement:** Find the derivative of the function $y = e^{4 \sin x} + \ln(2 + \sin x) + \cosh(\sin x)$. 2. **Recall formulas and rules:** - Derivative of $e^{u}$ is $e^{u} \cdot u'$. - Derivative of $\ln(u)$ is $\frac{u'}{u}$. - Derivative of $\cosh(u)$ is $\sinh(u) \cdot u'$. - Chain rule: derivative of composite function $f(g(x))$ is $f'(g(x)) \cdot g'(x)$. - Derivative of $\sin x$ is $\cos x$. 3. **Calculate derivatives step-by-step:** - For $e^{4 \sin x}$: Let $u = 4 \sin x$, then $u' = 4 \cos x$. So, $\frac{d}{dx} e^{4 \sin x} = e^{4 \sin x} \cdot 4 \cos x$. - For $\ln(2 + \sin x)$: Let $v = 2 + \sin x$, then $v' = \cos x$. So, $\frac{d}{dx} \ln(2 + \sin x) = \frac{\cos x}{2 + \sin x}$. - For $\cosh(\sin x)$: Let $w = \sin x$, then $w' = \cos x$. Derivative of $\cosh w = \sinh w \cdot w'$. So, $\frac{d}{dx} \cosh(\sin x) = \sinh(\sin x) \cdot \cos x$. 4. **Combine all parts:** $$ \frac{dy}{dx} = 4 e^{4 \sin x} \cos x + \frac{\cos x}{2 + \sin x} + \sinh(\sin x) \cos x $$ 5. **Final answer:** $$ \boxed{\frac{dy}{dx} = 4 e^{4 \sin x} \cos x + \frac{\cos x}{2 + \sin x} + \sinh(\sin x) \cos x} $$