# Thread: (Help) Pure Core 3 - Maths homework

1. 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 ):

Thanks!

2.

3. You used the formula for (uv)' but what you need is (u/v)' = (vu' - uv')/v^2.

4. Originally Posted by mathman
You used the formula for (uv)' but what you need is (u/v)' = (vu' - uv')/v^2.
The quotient rule is for two functions being divided, isn't it?

5. Originally Posted by JNicko
ai) Find dy/dx when y=(e^x)/sin2x
I think mathman's referring to this, unless that's a typo.

Assuming you've got part ai right, part aii looks fine. (The origin would be x=0.)

For part b, just do the same thing again, but use x=pi instead of x=0.

6. Originally Posted by JNicko
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)
No, this is not correct, and everything that follows is incorrect as well:

u=e^x v=sin2x
u'=e^x v'=2cos2x

Actually u'=e^x(v'+v) (not that it helps you in finding dy/dx)

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