How do you get the derivative of this?

  • Thread starter GrifteR150
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In summary, the problem involves finding the current i for a given equation q=5te^(-10^(3)t) Coulombs, using the formula i=dq/dt. The solution involves using the product rule and letting f=t and g=e^(-10^(3)t) to find g' by applying the chain rule. The given answer in the book is i=5(1-10^3t)e^(-10^3t) Amps. However, a small correction in the calculation of g' (g'=-10^3e^(-10^(3)t)) will result in the same answer as the book.
  • #1
GrifteR150
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This is for my Circuits class, I am supposed to find the current i.

If q = 5te^(-10^(3) t) Coulombs


Find i (current) the formula for i is: i = dq/dt


So what I did was try to find the derivative of q, which is dq


So I brought out the constant 5, and I am left with te^(-10^(3) t)

and I recognize this as a product of two functions so I tried the product rule.


I let my first function f = t, and my second function g = e^(-10^(3) t)


so f ' = 1 and g ' (using the chain rule) i got (e^(-10^(3) t) x '10^3)



so i plugged it into the product role (fg)' = fg' + gf' and i get a weird answer.


The given answer in the back of the book is :

i = 5(1-10^3 t)e^(-10^3 t) Amps

can someone help me please
 
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  • #2
Perhaps you just need to simplify? What does your current answer look like?

PS: we have a homework section :wink:
 
  • #3
Your general approach is correct. g' = e^(-10^(3) t) * -10^3

When you plug this into the product rule, you get a "weird answer?" Are you sure that your weird answer is not, in fact, the correct answer in unsimplified form?

- Warren
 
  • #4
It looks to me that if you make a small correction in your calculation of g', then you will get the same answer as the book.
 
  • #5
GrifteR150 said:
This is for my Circuits class, I am supposed to find the current i.

If q = 5te^(-10^(3) t) Coulombs


Find i (current) the formula for i is: i = dq/dt


So what I did was try to find the derivative of q, which is dq


So I brought out the constant 5, and I am left with te^(-10^(3) t)

and I recognize this as a product of two functions so I tried the product rule.


I let my first function f = t, and my second function g = e^(-10^(3) t)


so f ' = 1 and g ' (using the chain rule) i got (e^(-10^(3) t) x '10^3)
I presume that is a typo and it should be (e^(-10^(3)t) x (-10^3)) not
'10^3.

so i plugged it into the product role (fg)' = fg' + gf' and i get a weird answer.
How weird? You just said [itex]f= t, g'= -10^3 e^{-10^3t}, g= e^{-10^3t}, f'= 1[/itex] so [itex]fg'+ f'g= -10^3 te^{-10^3t}+ e^{-10^t}[/itex]
What do you get if you factor [itex]e^{-10^3 t}[/itex] out of that?

The given answer in the back of the book is :

i = 5(1-10^3 t)e^(-10^3 t) Amps

can someone help me please
 
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FAQ: How do you get the derivative of this?

How do you find the derivative of a function?

The derivative of a function is found by taking the limit of the slope of the tangent line at a specific point on the function. This is denoted by the notation f'(x) or dy/dx.

What is the purpose of finding the derivative?

The derivative is used to determine the rate of change of a function at a specific point, as well as the slope of the tangent line at that point. It is also used to find the maximum and minimum values of a function.

What is the difference between the derivative and the slope of a line?

The derivative is a general concept that can be applied to any function, while the slope of a line is only applicable to linear functions. The derivative represents the instantaneous rate of change at a specific point, while the slope of a line represents the average rate of change over a given interval.

How do you find the derivative of a composite function?

To find the derivative of a composite function, you can use the chain rule. This involves taking the derivative of the outer function and multiplying it by the derivative of the inner function.

Can the derivative of a function be negative?

Yes, the derivative of a function can be negative. This means that the function is decreasing at that specific point. A positive derivative indicates that the function is increasing at that point, and a derivative of zero indicates a horizontal tangent line.

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