MHB Cauchy-Riemann Equations test problem

[jaychay]

I cannot figure it out which choice is correct
Please help me
Please!
Thank you in advance

 

[I like Serena]

What is the real part of $f(x+iy)$? That is $u(x,y)$.
What is its partial derivative with respect to $x$? That is $u_x$.

 

[jaychay]

What is the real part of $f(x+iy)$? That is $u(x,y)$.
What is its partial derivative with respect to $x$? That is $u_x$.

Is the correct answer is choice D ?
I already done it by using partial derivatives and compare them

 

[I like Serena]

Is the correct answer is choice D ?
I already done it by using partial derivatives and compare them

It seems you are doing some online exam.
I'd rather not validate answers for something like that.

 

[jaychay]

It seems you are doing some online exam.
I'd rather not validate answers for something like that.

Sir, it's not an online exam
It's my assignment that I try to do on my own
I translate it from my math textbook in my country into English
I really want to become better at math.

If you still don't believe or you think I try to lie to you
Here is the original form of math textbook in my country

 

[Prove It]

Have you at least established what your u and v are?

As Klaas stated, $\displaystyle \begin{align*} u = \mathcal{Re}\left( f \right) = \mathrm{e}^x \left[ y + \cos{ \left( y \right) } \right] \end{align*}$ and $\displaystyle \begin{align*} v = \mathcal{Im}\left( f \right) = \mathrm{e}^x \sin{ \left( y \right) } \end{align*}$. Surely you can evaluate the necessary partial derivatives and make a judgement...

 

[jaychay]

Have you at least established what your u and v are?

As Klaas stated, $\displaystyle \begin{align*} u = \mathcal{Re}\left( f \right) = \mathrm{e}^x \left[ y + \cos{ \left( y \right) } \right] \end{align*}$ and $\displaystyle \begin{align*} v = \mathcal{Im}\left( f \right) = \mathrm{e}^x \sin{ \left( y \right) } \end{align*}$. Surely you can evaluate the necessary partial derivatives and make a judgement...

Thank you very much for guiding it for me.