Let \(w=x\sin y,\ x=e^t,\ y=\pi-t\)

Apply the chain Rule for One independet Variable

\(\frac{dw}{dt}=\frac{dw}{dx}\frac{dx}{dt}+\frac{dw}{dy}\frac{dy}{dt}\)

So

\(\frac{dw}{dt}=\frac{d}{dx}[x\sin y]\frac{d}{dt}[e^t]+\frac{d}{dy}[x\sin y]\frac{d}{dt}[\pi-t]\)

Calculate the Partial derivatives, you get

\(\frac{dw}{dt}=\sin y(e^t)+(x\cos y)(-1)\)

Simplify

\(\frac{dw}{dt}=e^t\sin y-x\cos y\)

Substitute \(x=e^t,\ y=\pi-t,\) so

\(\frac{dw}{dt}=e^t\sin(\pi-t)-e^t\cos(\pi-t)\)

Remember that \(\sin(\pi-t)=\sin t\) and \(\cos(\pi-t)=-\cos t ,\) so

\(\frac{dw}{dt}=e^t\sin t-e^t(-\cos t)\)

\(\frac{dw}{dt}=e^t(\sin t+\cos t)\)

Evaluate \(\frac{dw}{dt}\) when t =0

\(\frac{dw}{dt}=e^0(\sin0+\cos0)\)

\(\frac{dw}{dt}=1(0+1)\)

\(\frac{dw}{dt}=1\)

Results

\(e^t(\sin t+\cos t),\ 1\)

Apply the chain Rule for One independet Variable

\(\frac{dw}{dt}=\frac{dw}{dx}\frac{dx}{dt}+\frac{dw}{dy}\frac{dy}{dt}\)

So

\(\frac{dw}{dt}=\frac{d}{dx}[x\sin y]\frac{d}{dt}[e^t]+\frac{d}{dy}[x\sin y]\frac{d}{dt}[\pi-t]\)

Calculate the Partial derivatives, you get

\(\frac{dw}{dt}=\sin y(e^t)+(x\cos y)(-1)\)

Simplify

\(\frac{dw}{dt}=e^t\sin y-x\cos y\)

Substitute \(x=e^t,\ y=\pi-t,\) so

\(\frac{dw}{dt}=e^t\sin(\pi-t)-e^t\cos(\pi-t)\)

Remember that \(\sin(\pi-t)=\sin t\) and \(\cos(\pi-t)=-\cos t ,\) so

\(\frac{dw}{dt}=e^t\sin t-e^t(-\cos t)\)

\(\frac{dw}{dt}=e^t(\sin t+\cos t)\)

Evaluate \(\frac{dw}{dt}\) when t =0

\(\frac{dw}{dt}=e^0(\sin0+\cos0)\)

\(\frac{dw}{dt}=1(0+1)\)

\(\frac{dw}{dt}=1\)

Results

\(e^t(\sin t+\cos t),\ 1\)