- #1

- 35

- 0

- Thread starter gokuls
- Start date

- #1

- 35

- 0

- #2

member 392791

- #3

- 35

- 0

Okay, could you please explain the intuition?

- #4

- 811

- 6

h(x) = e^(2x)

f(x) = e^x

g(x) = x^2

Our the function we're interested in

(g ∘ f)(x) = g(f(x)) = (f(x))^2 = ((e^x)^2) = e^(2x) = h(x)

You know how to use the chain rule here. What you're asking is why can't we write it:

h(x) = e^(2x) = e^(x2) = (e^2)^x = ...

and go on from there to use the chain rule on different functions. (Let me know if this isn't what you meant).

The problem is doing that, you can't end up with the composition of two interesting functions. In order to decompose functions, you need all instances of "x" inside the parentheses. (This isn't a hard, fast law, but you need to be able to factor out a all instances of x).

The chain rule is the only thing that's going to help you with this one, I think.

- #5

- 811

- 6

What is the intuition that x^(yz) = (x^y)^z?

If you can do that one, perhaps it isn't so hard to understand why (g ∘ f)' = (g' ∘ f) f'

- #6

- 1,254

- 106

Have you worked out the other derivative rules using the limit definition?

- #7

Stephen Tashi

Science Advisor

- 7,476

- 1,416

The quotient in the definition of derivative resembles a magnifcation ratio. The size of the original image is (x+h) - h = h. The size of the magnified (or contracted) image is f(x + h) - f(x) The ratio of magnification for the "lens" f is (f(x+h)-f(x))/h. For small h, this is approximately the derivative f'(x).

The composition of two function f(g(x)) is thought about as the application of two lenses to the image. We send the original image through the lens g. Then we send that image though the lens f. Thinking about it this way, it is intuitive that the total magnification of the process is the product of the magnification of f times the magnifcation of g.

Think of e^(2x) as f(g(x)) with g(x) = 2x and f(t) = e^t, where t = g(x). The magnifcation of g is g'(x) = 2. The magnifcation of f is f'(t) = e^t where t = g(x). Think of x passing through both lenses. It goes into g first and then into f. The lens g magnifes by 2. The lens f magnifies by e^t = e^(g(x) = e^(2x). The total magnification is the product (2)(e^(2x)).

You can look at it that way by using f(t) = t^2 and g(x) = e^x. The magnification of f is f'(t) = 2t. The magnifcation of g is g'(x) = e^x. The product is (2t)(e^x) = (2 g(x) ) e^x = (2 e^x) (e^x) = 2 e^(2x).At any given point the function e^(2x) is squared the original function of e^x.

- #8

- 647

- 3

[tex]\left(\left(e^x\right)^2\right)'\ne\left(\left(e^x\right)'\right)^2[/tex]

The reason is simply that

[tex]\left(f^2\left(x\right)\right)' \ne\left(f'\left(x\right)\right)^2[/tex]

Use the product rule if you want to see why this is.

- #9

- 300

- 22

Now instead of e^x consider e^(2x). When 2x is around A, the functions have the same value, but if you now increase x by 0.01, the whole exponent increases by 0.02, which as we saw before, increasing the exponent by 0.02 increases the function by 0.08, so we get twice as much change in the function's value per an increase in x of 0.01 as we did before, because the exponent increases twice as much as before per the same change in x.

The exponent doesn't have to be linear like that, you just need to know how much does the whole exponent increase with changes of x and that is what the derivative of the exponent shows.

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 29K

- Last Post

- Replies
- 9

- Views
- 855

- Last Post

- Replies
- 2

- Views
- 647

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 16

- Views
- 17K

- Last Post

- Replies
- 17

- Views
- 14K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 18

- Views
- 10K

- Replies
- 5

- Views
- 6K