How to do this kind of differentation help?

Question

How do you get
d/dx (e^x)

Sorry, a 15-yr-old mind cannot easily absorb this kind of differentiation!! (I'm only a 4th yr student)

Thanks in advance!!!
^_^

I know that
d/dx (v^n) = nv^(n - 1)dv/dx but can this help in my problem? (I know something about getting derivatives, integrals, rate of change, etc.)
^_^

Answer 1

Use these facts: ln(x) is the natural logarithm of x and so
ln(e^x) = x and d/dx ln(x) = 1/x

To prove d/dx e^x = e^x you find the derivative in two ways, and equate them:

First, trivially, since ln(e^x) = x it follows that:

d/dx ln(e^x) = dx/dx = 1

But also, if we let u=e^x, then by the chain rule:

d/dx ln(e^x) = d/dx ln(u) = d/du log(u)*du/dx =

1/u du/dx = 1/e^x d/dx e^x (since u = e^x)

Equating these 2 results we have:

1/e^x d/dx e^x = 1 and multiplying through by e^x we have

d/dx e^x = e^x

Hope that's clear enough. You are doing very well for your age. Keep up the good work!

Answer 2

I'm not perfectly sure what you're asking, but I'll try anyway.

dy/dx is called differential notation. At your level of Calculus, it's just another way of writing f'(x). f'(x) is functional notation.

In answer to your questions, the derivative of e^x is also e^x. It has the very nice property that dy = y. (The differential is equal to the original equation.)

Answer 3

To solve this, remember that to differentiate e to the power x, one get's the derivative of the power of x and then multiplies it by the equation itself i.e, Equation is = e^x
Find derivative of x, which is = 1, (i.e. d/dx of x^1 = 1)
(cause the power of x is 1, hence we reduce the power by 1
when differentiating and then multiply it by what remains).
We then multiply the derivative by the equation itself, i.e.,
1 * e^x = e^x, therefore differentiating e^x gives, = e^x

Hope I assisted you - hard to explain via the net, but all the best.

Answer 4

this could be a appropriate expenditures subject. Draw a diagram as you describe [ a incredible angled triangle with hyp =9 base = x and height = y = sqrt[ 80 one - x^2] all of us comprehend dx/dt = 3 and dy/dt = dy/dx * dx/dt so locate dy/dx from y = sqrt[ 80 one - x^2] use y = 7.8

Answer 5

look i think you don't know the series of e^x;
e^x=summation(x^n/n!) where n varies from 0 to infinity
try to write this and then diffrentiate the terms individually using the formula that dy/dx (x^n)= n x^(n-1) . you will notice that every term becomes the earlier term upon differentiation.

feel free to contact me thru email.

Answer 6

Remember there is a formula that dy/dx of e^x is e^x.

Answer 7

i dont know how to explain it either, but:

d/dx {e^f(x)} = f'(x) e^f(x) where f'(x) is the differential of f(x)

so,

d/dx e^x = d(x)/dx e^(x)
= (1) e^x
= e^x

Answer 8

d/dx (e^x) = e^x.

It's one of the few functions whose derivative is equal to itself.

Answer 9

The derivate of e^x is e^x.
e^x = exp(x)
It is the solution of y'= y at any x.

Only this functions shows this trick.
Good luck to you.

Answer 10

f(x) = e^x

df(x)/dx = e^x