Let $g:\mathbb{R} \rightarrow \mathbb{R}$ continuous and $f:\mathbb{R}^{3} \rightarrow \mathbb{R}$ defined for $f(x,y,z)=\int_{\sin x}^{yz} g(t)dt$. Express $f$ like the composition of differentiable functions and use the chain's rule to find $f'(x,y,z)$,
I think the functions are $H(x,y,z) = (\sin x,y,z)$, $G(x,y,z)= \int_{x}^{yz} g(t)dt$, so $f(x,y,z)= G \circ H$. Did I choose the correct functions? Can you help me with the chain's rule?
Sure, that's true, but not very useful (if this is your first time seeing this stuff). Instead, consider \begin{align} F(\xi):= \int_0^{\xi}g(t)\,dt. \end{align} Then, \begin{align} f(x,y,z)&=\int_0^{yz}g(t)\,dt - \int_0^{\sin x}g(t)\,dt\\ &=F(yz)-F(\sin x) \end{align} So, define $\phi_1(x,y,z)=yz$ and $\phi_2(x,y,z)=\sin x$. Then, \begin{align} f(x,y,z)&=(F\circ\phi_1)(x,y,z)-(F\circ \phi_2)(x,y,z) \end{align} So, what does the chain rule tell you now? Note that for differentiating $F$, the fundamental theorem of calculus is useful (assuming of course $g$ is continuous).