Find the gradient of the tangent to the curve y = x sin x^2 at the point where x = π/3
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 $\displaystyle x=\frac{\pi}3$
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
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 .