How do you differentiate $y = sin (arccos t)$ $y=sin(arccost)$ ?

Question

1804 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-18T09:41:24

We will use the chain rule.

$y = sin (arccos t)$ $y=sin(arccost)$

so

$\frac{d y}{d t} = cos (arccos t) \cdot \frac{d (arccos t)}{d t} =$ $dy/dt=cos(arccost)*(d(arccost))/dt=$

$cos (arccos t) (- \frac{1}{\sqrt{1 - t^{2}}}) = - \frac{cos (arccos t)}{\sqrt{1 - t^{2}}}$ $cos(arccost)(-1/sqrt(1-t^2))=-cos(arccost)/sqrt(1-t^2)$

so

$\frac{d y}{d t} = - \frac{cos (arccos t)}{\sqrt{1 - t^{2}}}$ $dy/dt=-cos(arccost)/sqrt(1-t^2)$

How do you differentiate y=sin(arccost)y=sin(arccost)y=sin(arccost)?