# Find the derivative of a function

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1. Sep 13, 2016

### ChrisBrandsborg

1. The problem statement, all variables and given/known data
E(t) = 100te1+cos((2π*t)/365

What is the derivative of this function?

2. Relevant equations

3. The attempt at a solution

100*e1+cos((2π*t)/365) + 100t*e1+cos((2π*t)/365) * -(2π/365)sin(2πt/365)

I have tried to use the rules for derivative of products, and also used the chain rule.

Last edited: Sep 13, 2016
2. Sep 13, 2016

### Simon Bridge

The sine should not be inside the exponent - example:
$\frac{d}{dt}e^{\sin(t)} = e^{\sin(t)}\cos(t)$

Note:
It can make for better formatting to use $\exp[f(t)]$ instead of $e^{f(t)}$ when $f(t)$ is quite involved.

3. Sep 13, 2016

### ChrisBrandsborg

But other than that, is it correct? So can I put:

100e1+cos(2πt)/365) (1 - (2π/365)sin(2πt/365) t) ?

4. Sep 13, 2016

### Simon Bridge

$\frac{d}{dt}\big[1+\cos(2\pi t/365) \big] \neq 1-(2\pi/365)\sin(2\pi t/365)$
The main trick with using chain rule is to keep careful track of what is what.

Try $g = e^{u(v(t))}$ so $\frac{dg}{dt} = e^{u(v(t))}\frac{du}{dv}\frac{dv}{dt}$ and work out du/dv and dv/dt separately.

[no - got it right first time ...]

Last edited: Sep 13, 2016
5. Sep 13, 2016

### vela

Staff Emeritus
That's not what he's doing.

This is correct, though it's more common to put the factor of $t$ in front of the sin. That way there's no ambiguity it's multiplying the function rather than the argument of the function.

6. Sep 13, 2016

### Simon Bridge

Oh I'm having fun reading the notation ... like reading "*-" as a "-" instead of "multiplied by the negative of". My bad:
$$\frac{d}{dt}E(t) = 100\left[1-(2\pi /365)t\sin(2\pi t/365) \right]e^{1+\cos(2\pi t/365)}$$ ... something like that?.

7. Sep 13, 2016

### ChrisBrandsborg

Yes, that is what I have now. Is it done? Or can I simplify it? :)

8. Sep 13, 2016

### Simon Bridge

I'd stop there.