F(x) = e^(2x) and g(x) = lnx. what is the derivative of f(g(x)) at x=e

  • Thread starter meredith
  • Start date
  • Tags
    Derivative
In summary, the derivative of f(x) = e^(2x) is f'(x) = 2e^(2x). The derivative of g(x) = lnx is g'(x) = 1/x. The composite function f(g(x)) is e^(2lnx). The derivative of f(g(x)) at x=e is 2e^2, which is found by using the chain rule and substituting x=e into f'(g(e)) and g'(e).
  • #1
meredith
16
0

Homework Statement


f(x) = e^(2x) and g(x) = lnx. what is the derivative of f(g(x)) at x=e


Homework Equations


dy/dx = f'g(x) x g'(x)


The Attempt at a Solution


i can't figure it out!
 
Physics news on Phys.org
  • #2
Try it! What's f'(x)? g'(x)? f'(g(x))?
 
  • #3
remember your chain rule:
[tex][f(g(x))]'=f'(g(x))\cdot g'(x)[/tex]
take each derivative separately and put it all together if it seems too complicated, that helps me a lot
 

Related to F(x) = e^(2x) and g(x) = lnx. what is the derivative of f(g(x)) at x=e

1. What is the derivative of f(x) = e^(2x)?

The derivative of f(x) = e^(2x) is f'(x) = 2e^(2x).

2. What is the derivative of g(x) = lnx?

The derivative of g(x) = lnx is g'(x) = 1/x.

3. What is the composite function f(g(x))?

The composite function f(g(x)) is e^(2lnx).

4. What is the derivative of f(g(x)) at x=e?

The derivative of f(g(x)) at x=e is f'(g(e)) * g'(e) = (2e^(2lnx))|x=e * (1/x)|x=e = 2e^2.

5. How can the chain rule be applied to find the derivative of f(g(x)) at x=e?

The chain rule can be applied by first finding the derivative of f(g(x)) and then substituting x=e into both f'(g(e)) and g'(e).

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
930
  • Calculus and Beyond Homework Help
Replies
2
Views
276
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
627
  • Calculus and Beyond Homework Help
Replies
8
Views
534
  • Calculus and Beyond Homework Help
Replies
2
Views
423
  • Calculus and Beyond Homework Help
Replies
2
Views
980
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
793
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Back
Top