Question

find the derivative of the function.
f(θ) = cos(θ²)
f(θ) =

1. - / 1 points

find the derivative of the function.
f(t) = 9t sin(πt)
f(t) =

Explanation:

Step1: Apply chain - rule for $f(\theta)=\cos(\theta^{2})$

Let $u = \theta^{2}$, then $f(u)=\cos(u)$. The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$, and the derivative of $u=\theta^{2}$ with respect to $\theta$ is $2\theta$. By the chain - rule $\frac{df}{d\theta}=\frac{df}{du}\cdot\frac{du}{d\theta}$, so $f'(\theta)=-\sin(\theta^{2})\cdot2\theta=- 2\theta\sin(\theta^{2})$.

Step2: Apply product - rule for $f(t)=9t\sin(\pi t)$

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $t$, then $y'=u'v + uv'$. Let $u = 9t$ and $v=\sin(\pi t)$. The derivative of $u = 9t$ with respect to $t$ is $u'=9$. The derivative of $v=\sin(\pi t)$ with respect to $t$ is $v'=\pi\cos(\pi t)$ (using the chain - rule with $w=\pi t$, derivative of $\sin(w)$ with respect to $w$ is $\cos(w)$ and derivative of $w=\pi t$ with respect to $t$ is $\pi$). Then $f'(t)=9\sin(\pi t)+9t\cdot\pi\cos(\pi t)=9\sin(\pi t)+9\pi t\cos(\pi t)$.

Answer:

$f'(\theta)=-2\theta\sin(\theta^{2})$
$f'(t)=9\sin(\pi t)+9\pi t\cos(\pi t)$