Hey everyone

Just a little stuck on some maths work

No, I didn't create an account just to post this, I forgot my old accounts details ):

So yeah, I'd appreciate it if someone could give me a few hints?

ai) Find dy/dx when y=(e^x)/sin2x

u=e^x v=sin2x

u'=e^x v'=2cos2x

((e^x)*2cos2x) + ((e^x)*sin2x) = (e^x)(sin2x+2cos2x)

Now for what i'm stuck on...

aii) Hence find the equation of the tangent to the curve y=(e^x)sin2x at the origin.

So, do I let x=0, since it says at the origin?

It doesn't imply any other x value

If so, (e^0)(sin0+2cos0) = (1)(2) = 2, so the gradient of the curve at x=0 is 2..

To find y, input x=0 into the original.. (e^0)(sin0) = (1)(0) = 0

So the tangent is y = 2x ?

Onwards to the next part

b) Show that the equation of the normal to the curve y=(e^x)sin2x at the point where x= pi is 2(e^pi)+x=pi

I haven't got a clue on how to do this one ):

So yeah, some proof reading and help please?

Thanks!