Differentiating $90 + 45 \cos(t^2/18)$ using the chain rule? Can someone please explain to me how to differentiate this function?
$$90 + 45 \cos(t^2/18)$$
I know you use the chain rule, but I can't get the final result correct.
 A: $$\zeta(t)=90+45\cos\left(\frac{t^2}{18}\right)$$
$$\require{cancel}\begin{align} \frac{d\zeta}{dt}&=\cancelto{0}{\frac{d}{dt}(90)}+\frac{d}{dt}\left(45\cos\left(\frac{t^2}{18}\right)\right)\\&= \frac{d}{dt}\left(45\cos\left(\frac{t^2}{18}\right)\right) \\&=45\left(-\sin\left(\frac{t^2}{18}\right)\right)\underbrace{\left(\frac{d}{dt}\left(\frac{t^2}{18}\right)\right)}_{\frac{t}{9}}\\& =-5t\sin\left(\frac{t^2}{18}\right)\end{align}$$
Because:
$$\frac{d}{dx}\cos(\varphi)=-\sin(\varphi)\cdot\frac{d\varphi}{dx}$$
A: Here by using the chain rule:
$\frac{d}{dt}(90+45\cos(\frac{t^2}{18})=\frac{d}{dt}(90)+45\frac{d}{dt}(\frac{t^2}{18})\cos'(\frac{t^2}{18})=0-45\frac{2t}{18}\sin(\frac{t^2}{18})=-5t\sin(\frac{t^2}{18}).$
A: Well the $90$ becomes simply zero. So you have
$$45\left(f\left(g(t)\right)\right)'$$
where $g(t)=t^2/18$ and cos $f(x)=\cos(x)$. The chain rule goes:
$$\left(f\left(g(t)\right)\right)'=f'(g(t))\cdot g'(t)$$
And $f'(x)=-\sin(x)$ so $f'(t^2/18)=-\sin(t^2/18)$.
and $g'(t)=t/9$, that's clear, right?
So
$$45\cdot f'(g(t))\cdot g'(t)=-45\cdot\sin(t^2/18)\cdot t/9$$
A: $\frac{d}{dt}90 = 0$ so we concentrate on your second term.
$\displaystyle\frac{d}{dt} \cos(t^2/18) = \frac{dt^2}{dt}\frac{d}{dt^2} \cos(t^2/18) = 2t \times (-\frac{1}{18}\sin(t^2/18))= -\frac{1}{9}t\sin(t^2/18)$
And then multiply by 45.
When you take the derivative of the $\cos()$, think of $t^2$ as a variable in its own right (which it really is, after all).  If it helps, replace it with another symbol, like $t^2 = u$, and then, when you have your result, put the $t^2$ back in where you have $u$.
A: Remark that $cos'(t)=-sin( t)$ and $\frac{d}{dt}(\frac{t^2}{18})=\frac{t}{9}$. Then
$$
\frac{d}{dt}(90+45cos(\frac{t^2}{18}))=\frac{d}{dt}(90)+\frac{d}{dt}(45cos(\frac{t^2}{18}))=0+45\frac{d}{dt}(cos(\frac{t^2}{18}))
$$
But setting $s=s(t)=\frac{t^2}{18}$, 
$$
\frac{d}{dt}(cos(\frac{t^2}{18}))=\frac{d}{ds}(cos(s))\frac{d}{dt}(s(t))=-sin(s).s'(t)=-sin(\frac{t^2}{18})\frac{t}{9}.
$$
Thus
$$
\frac{d}{dt}(90+45cos(\frac{t^2}{18}))=45(-sin(\frac{t^2}{18})\frac{t}{9})=-5t.sin(\frac{t^2}{18})
$$
A: You need to see the nesting:
$$
f(t) = 90 + 45\cos\left(\frac{t^2}{18}\right) \\
g(x) = 90 + 45\cos(x) \rightarrow g\left(x = \frac{t^2}{18}\right) = f(t) = 90 + 45\cos\left(\frac{t^2}{18}\right) \\
g'(x) = 90 - 45\sin(x) \rightarrow g'\left(x = \frac{t^2}{18}\right) = 90 - 45\sin\left(\frac{t^2}{18}\right) \\
x'(t) = \frac{d}{dt}\left(\frac{t^2}{18}\right) = \frac{t}{9} \\
\left.\left.\frac{d}{dt}\right(f(t)\right) = \left.\left.\frac{d}{dt}\right (g\left(x(t)\right)\right) = \frac{dx}{dt}\cdot\left(\frac{dg}{dx} \circ x(t)\right) \\
\frac{df}{dt} = \frac{t}{9}\left(90 - 45\sin\left(\frac{t^2}{18}\right)\right)
$$
