Find the tangent to the curve

Question

The equation of the tangent to the curve $y=sin²(\frac{\pi x^3}{6})$ at $x=1$ is? I know the question is pretty simple and straight but I would like to cross check my answer which is $y=\frac14 + \frac{\pi\sqrt3}{4}(x-1)$

Answer 1

I'm getting a different value for the slope. When taking your derivative, did you apply the chain rule twice? We have $y=u^2$, with $u=\sin(v)$ and $v=\frac{\pi x^3}{6}$. To find $\frac{dy}{dx}$, you need to multiply $\frac{dy}{du}$, $\frac{du}{dv}$ and $\frac{dv}{dx}$.

Answer 2

$$2y=1-\cos\dfrac{\pi x^2}3$$

$$2y'=\sin\dfrac{\pi x^2}3\cdot\dfrac{2\pi x}3$$

Answer 3

Take $f(1)$ for the intercept

$$f(1)=\frac{1}{4}=b$$

Take the derivative and evaluate it at $x=1$ for the slope of the line

$$f(x)'=2\sin\left(\frac{\pi x^3}{6}\right)\cos\left(\frac{\pi x^3}{6}\right)\frac{\pi x^2}{2}$$

$$f'(1)=\pi\frac{1}{2}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}\pi}{4}=m$$

Then the tangent line follows $y=m(x-1)+b$

Answer 4

You are indeed correct.

Let $y=\sin^2[{f(x)}]$

Then $y=\frac12-\frac12\cos[2f(x)]$

$\to\frac{dy}{dx}=f'(x)\sin[2f(x)]$

$f(x)=\frac\pi6x^3\to f'(x)=\frac\pi2x^2$

$\to \frac{dy}{dx}=\frac{\pi}{2}x^2\sin(\frac{\pi x^3}{3})$

Hence $x=1\to \frac{dy}{dx}=\frac\pi2\sin(\frac\pi3)=\frac{\pi\sqrt3}{4}$

$x=1\to y=\frac14, y-\frac14=\frac{\pi\sqrt3}{4}(x-1)\to y=\frac14+\frac{\pi\sqrt3}{4}(x-1)$