Find the derivative of a function

In summary: You could put the factor of 100 in with the ##\exp(1+\cos(2\pi t/365)## if you want.In summary, the derivative of the function E(t) = 100te1+cos((2π*t)/365 is 100(1-(2π/365)t*sin(2πt/365))e1+cos(2πt/365).
  • #1
ChrisBrandsborg
120
1

Homework Statement


E(t) = 100te1+cos((2π*t)/365

What is the derivative of this function?

Homework Equations

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.
 
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  • #2
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
Simon Bridge said:
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.
But other than that, is it correct? So can I put:

100e1+cos(2πt)/365) (1 - (2π/365)sin(2πt/365) t) ?
 
  • #4
##\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 ...]
 
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  • #5
Simon Bridge said:
##\frac{d}{dt}\big[1+\cos(2\pi t/365) \big] \neq 1-(2\pi/365)\sin(2\pi t/365)##
That's not what he's doing.

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

100e1+cos(2πt)/365) (1 - (2π/365)sin(2πt/365) t) ?
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
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
Simon Bridge said:
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?.

Yes, that is what I have now. Is it done? Or can I simplify it? :)
 
  • #8
I'd stop there.
 

FAQ: Find the derivative of a function

1. What is the definition of a derivative?

The derivative of a function is the rate of change of the function at a specific point. In other words, it measures how much the function is changing at that point.

2. How do you find the derivative of a function?

To find the derivative of a function, you can use the derivative rules such as the power rule, product rule, quotient rule, and chain rule. These rules allow you to find the derivative of a function by manipulating the original function.

3. What is the purpose of finding the derivative of a function?

The derivative of a function is useful in many applications, such as optimization, finding maximum and minimum points, and determining the slope of a tangent line at a specific point on a curve. It also helps in understanding the behavior of a function and its rate of change.

4. Can you find the derivative of any type of function?

Yes, the derivative can be found for any continuous function. However, some functions may require more advanced techniques to find the derivative, such as the chain rule for composite functions.

5. How can I check if I have correctly found the derivative of a function?

You can check your derivative by graphing both the original function and its derivative on the same graph. The derivative should be a tangent line to the original function at every point. Additionally, you can use the first principles definition of a derivative to verify your answer.

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