# Pretty easy trig. question

• Jul 20th 2008, 07:49 PM
sweetG
Pretty easy trig. question
Hi I've already posted this question up although its actually getting pretty urgent that I finish it because its due very soon so could someone please help me. Its pretty easy but for some reason I just can't think straight today

Find the gradient of the tangent to the curve y = x sin x^2 at the point where x = π/3

I've alreaf '(x) = sinx^2 + x(2x cos x^2)dy done the first bit and found the gradient which is:

f '(x) = sinx^2 + x(2x cos x^2)

but for some strange reason I'm real confused about where to substitute x = π/3 because I see a lot of x's there but I'm not too sure on which ones I should sub this in for. I know that this should be very easy but I just can't think straight today >_<
• Jul 20th 2008, 07:52 PM
kalagota
..sub it to all x's that you see.. :)
• Jul 20th 2008, 08:01 PM
sweetG
f '(x) = sinx^2 + x(2x cos x^2)

So would it be:

When x =π/3

f '(x) = sin(π^2/9) + π/3(2π/3 cos π^2/9)

Is that correct ?

Or have I done something wrong ?
• Jul 20th 2008, 08:22 PM
kalagota
yes..