# (Help) Pure Core 3 - Maths homework

• October 7th, 2014, 06:36 PM
JNicko
(Help) Pure Core 3 - Maths homework
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! :D
• October 7th, 2014, 06:54 PM
mathman
You used the formula for (uv)' but what you need is (u/v)' = (vu' - uv')/v^2.
• October 7th, 2014, 06:58 PM
JNicko
Quote:

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?
• October 7th, 2014, 09:05 PM
MagiMaster
Quote:

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.
• October 8th, 2014, 12:18 AM
Howard Roark
Quote:

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: