# 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.

-

$$\zeta(t)=90+45\cos\left(\frac{t^2}{18}\right)$$ \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: $$\frac{d}{dx}\cos(\varphi)=-\sin(\varphi)\cdot\frac{d\varphi}{dx}$$

-
thank you this made so much sense! Splitting up the equation helped me see what i had to differentiate. The 1/9 was were I went wrong and messed up. Thanks!!! –  sloth1111 May 7 '14 at 0:24
@sloth1111 No problem. Good luck. –  Shahar May 7 '14 at 0:25

Here by using the chain rule:

$\frac{d}{dt}(90+45\cos(\frac{t^2}{18})=\frac{d}{dt}(90)+45\frac{d}{dt}(\frac{t^2}{18})\cos'(\frac{t^2}{18})=0-45\frac{2t}{18}\sin(\frac{t^2}{18})=-5t\sin(\frac{t^2}{18}).$

-

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 $$45\cdot f'(g(t))\cdot g'(t)=-45\cdot\sin(t^2/18)\cdot t/9$$

-

$\frac{d}{dt}90 = 0$ 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 $u$.

-

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 $$\frac{d}{dt}(90+45cos(\frac{t^2}{18}))=45(-sin(\frac{t^2}{18})\frac{t}{9})=-5t.sin(\frac{t^2}{18})$$

-

You need to see the nesting:

$$f(t) = 90 + 45\cos\left(\frac{t^2}{18}\right) \\ g(x) = 90 + 45\cos(x) \rightarrow g\left(x = \frac{t^2}{18}\right) = f(t) = 90 + 45\cos\left(\frac{t^2}{18}\right) \\ g'(x) = 90 - 45\sin(x) \rightarrow g'\left(x = \frac{t^2}{18}\right) = 90 - 45\sin\left(\frac{t^2}{18}\right) \\ x'(t) = \frac{d}{dt}\left(\frac{t^2}{18}\right) = \frac{t}{9} \\ \left.\left.\frac{d}{dt}\right(f(t)\right) = \left.\left.\frac{d}{dt}\right (g\left(x(t)\right)\right) = \frac{dx}{dt}\cdot\left(\frac{dg}{dx} \circ x(t)\right) \\ \frac{df}{dt} = \frac{t}{9}\left(90 - 45\sin\left(\frac{t^2}{18}\right)\right)$$

-