Derivative of e^(kx) without using the Chain Rule

[apple2357]

Assuming we have established the derivative of e^x is e^x, can someone offer me an intuitive way of seeing why e^(kx) differentiates to ke^(kx).
I know its possible to use the Chain Rule, but i want to avoid that. I feel there is something graphical, but i can't quite grasp it.

 

[fresh_42]

Well, you could use the Taylor series instead.

What is [imath] y=e^{kx}? [/imath] It is the solution to the differential equation [imath] y'=k \cdot y [/imath] with [imath] y(0)=1 [/imath] and [imath] y'(0)=k. [/imath]
Here is an article about that relationship: https://www.physicsforums.com/insig...onship-between-integration-and-eulers-number/

You can view this question from many perspectives, but in the end, you have to use something. E.g., what is [imath] e [/imath] in your question?

 

[Dr.Peterson]

Assuming we have established the derivative of e^x is e^x, can someone offer me an intuitive way of seeing why e^(kx) differentiates to ke^(kx).
I know its possible to use the Chain Rule, but i want to avoid that. I feel there is something graphical, but i can't quite grasp it.

If you're looking for an intuitive understanding rather than a proof, you might consider that e^(kx) is just e^x compressed horizontally by a factor of 1/k (since x has to be 1/k as large to obtain the same y). That will multiply the slope by k; think about it if that is not clear. And, of course, draw the graphs (including sample tangent lines) to visualize it.

 

[apple2357]

If you're looking for an intuitive understanding rather than a proof, you might consider that e^(kx) is just e^x compressed horizontally by a factor of 1/k (since x has to be 1/k as large to obtain the same y). That will multiply the slope by k; think about it if that is not clear. And, of course, draw the graphs (including sample tangent lines) to visualize it.

Thank you, i think thats exactly what i was after.