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

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- Jul 20th 2008, 08:29 AMsweetGThis is a pretty straight foward question (Trig.)
Find the gradient of the tangent to the curve y = x sin x^2 at the point where x = π/3

- Jul 20th 2008, 08:34 AMMoo
Hello,

Let $\displaystyle f(x)=x \sin x^2$

The gradient of the tangent at the point where $\displaystyle x=\frac{\pi}{3}$ is $\displaystyle f'(\pi/3)$.

So first of all, calculate $\displaystyle f'(x)$ (applying product rule & chain rule) and then take $\displaystyle x=\frac{\pi}3$ - Jul 20th 2008, 08:54 AMsweetG
Umm I'm really bad at differentiating trigonometric functions but could the answer possibly be: f '(x) = sinx^2 + 2x cos x^2

- Jul 20th 2008, 08:58 AMMoo
- Jul 20th 2008, 09:08 AMsweetG
f '(x) = sinx^2 + x(2x cos x^2)

Would that be the correct answer ?

And well if it is correct does that mean I just substitute x = π/3 into all the x's ?

Oh I'm so bad at this >_< - Jul 20th 2008, 09:29 AMMoo
- Jul 20th 2008, 09:56 AMsweetG
Yup it was a careless mistake

but am this is the part that confuses me as well

coz am i supposed to substitute it into every x value there ?

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

so when x = π/3

Am I supposed to put x = π/3 in to all x's coz I dunno if its just me but that looks wrong ?! =S I'm so confused

Where do I sub in π/3 >_< sorry its pretty late here and I'm just not thinking properly at the moment - Jul 20th 2008, 06:59 PMsweetG
I'm just saying that well if I substituted π/3 in every single x I saw then that would be wrong, right ?

Argh I hate trigonometry >_< - Jul 21st 2008, 05:32 AMtutor
Hi,

Given function is y=xsin(x^2)

to find the slope=m=y'=differentiate with respect to x by using product rule formula

y'=sinx^2+xcosx^2.2x

y'=sinx^2+2x^2cosx^2

gradient of the give curve at the point x=pi/3

y'=sin(pi/3)^2+2(pi/3)^3cos(pi/3)^2

here u can use calculator to get the answer .