12. -/1 points find y and y. y = cos(sin(9t)) y = y = resources read it

Explanation:

Step1: Apply chain - rule for first derivative

Let $u = \sin(9t)$ and $y=\cos(u)$. The chain - rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. We know that $\frac{dy}{du}=-\sin(u)$ and $\frac{du}{dt}=9\cos(9t)$. Substituting $u = \sin(9t)$ back in, we get $y'=-9\cos(9t)\sin(\sin(9t))$.

Step2: Apply product - rule and chain - rule for second derivative

The product - rule is $(uv)' = u'v + uv'$, where $u=-9\cos(9t)$ and $v = \sin(\sin(9t))$. $u'=81\sin(9t)$ and for $v$, let $w=\sin(9t)$, then $v = \sin(w)$ and $\frac{dv}{dt}=\frac{dv}{dw}\cdot\frac{dw}{dt}=9\cos(9t)\cos(\sin(9t))$. $y''=81\sin(9t)\sin(\sin(9t))-81\cos^{2}(9t)\cos(\sin(9t))$.

Answer:

$y'=-9\cos(9t)\sin(\sin(9t))$ $y''=81\sin(9t)\sin(\sin(9t)) - 81\cos^{2}(9t)\cos(\sin(9t))$