Find dy/dt y=ccos(t)+t^2sin(t)

$y = c \cos (t) + t^{2} \sin (t)$

Differentiate both sides of the equation.

$\frac{d}{d t} (y) = \frac{d}{d t} (c \cos (t) + t^{2} \sin (t))$

The derivative of $y$ with respect to $t$ is $y'$ .

$y'$

Differentiate the right side of the equation.

By the Sum Rule, the derivative of $c \cos (t) + t^{2} \sin (t)$ with respect to $t$ is $\frac{d}{d t} [c \cos (t)] + \frac{d}{d t} [t^{2} \sin (t)]$ .

$\frac{d}{d t} [c \cos (t)] + \frac{d}{d t} [t^{2} \sin (t)]$

Evaluate $\frac{d}{d t} [c \cos (t)]$ .

Since $c$ is constant with respect to $t$ , the derivative of $c \cos (t)$ with respect to $t$ is $c \frac{d}{d t} [\cos (t)]$ .

$c \frac{d}{d t} [\cos (t)] + \frac{d}{d t} [t^{2} \sin (t)]$

The derivative of $\cos (t)$ with respect to $t$ is $- \sin (t)$ .

$c (- \sin (t)) + \frac{d}{d t} [t^{2} \sin (t)]$

Evaluate $\frac{d}{d t} [t^{2} \sin (t)]$ .

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t^{2}$ and $g (t) = \sin (t)$ .

$c (- \sin (t)) + t^{2} \frac{d}{d t} [\sin (t)] + \sin (t) \frac{d}{d t} [t^{2}]$

The derivative of $\sin (t)$ with respect to $t$ is $\cos (t)$ .

$c (- \sin (t)) + t^{2} \cos (t) + \sin (t) \frac{d}{d t} [t^{2}]$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$c (- \sin (t)) + t^{2} \cos (t) + \sin (t) (2 t)$

Reorder terms.

$- c \sin (t) + t^{2} \cos (t) + 2 t \sin (t)$

Reform the equation by setting the left side equal to the right side.

$y' = - c \sin (t) + t^{2} \cos (t) + 2 t \sin (t)$

Replace $y'$ with $\frac{d y}{d t}$ .

$\frac{d y}{d t} = - c \sin (t) + t^{2} \cos (t) + 2 t \sin (t)$

Find dy/dt y=ccos(t)+t^2sin(t)

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