7. let ( f(x,y)=xysin(2y) ) with ( y = e^{x}+x ). calculate ( \frac{df}{dx} ).\na. ( ysin(2y)+x(sin(2y)+2ycos…

Explanation:

Step1: Apply the product rule

The product rule states that if (f(x,y) = u(x,y)v(x,y)), then (\frac{df}{dx}=\frac{\partial u}{\partial x}v + u\frac{\partial v}{\partial x}). Here (u = x) and (v=y\sin(2y)). So (\frac{\partial u}{\partial x}=1), and (\frac{df}{dx}=y\sin(2y)+x\frac{d}{dx}(y\sin(2y))).

Step2: Apply the chain - rule for (\frac{d}{dx}(y\sin(2y)))

The chain rule: if (z = y\sin(2y)), then (\frac{dz}{dx}=\frac{\partial z}{\partial y}\frac{dy}{dx}). First, find (\frac{\partial z}{\partial y}): (\frac{\partial z}{\partial y}=\sin(2y)+y\times2\cos(2y)=\sin(2y) + 2y\cos(2y)). Second, since (y = e^{x}+x), (\frac{dy}{dx}=e^{x}+1). Then (\frac{d}{dx}(y\sin(2y))=(\sin(2y)+2y\cos(2y))(e^{x}+1)).

Step3: Combine the results

Substitute (\frac{d}{dx}(y\sin(2y))) back into the expression from Step1: (\frac{df}{dx}=y\sin(2y)+x(\sin(2y)+2y\cos(2y))(e^{x}+1)).

Answer:

A. (y\sin(2y)+x(\sin(2y)+2y\cos(2y))(e^{x}+1))