Question

1. let ( f(x,y)=xysin(2y) ) with ( y = e^{x}+x ). calculate ( \frac{df}{dx} ).

a. ( ysin(2y)+x(sin(2y)+2ycos(2y))(e^{x}+1) ).
b. ( ysin(2y) ).
c. ( x(sin(2y)+2ycos(2y)) ).
d. ( ysin(2y)+x(sin(2y)+2ycos(2y)) ).

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