Derivative of e^(2x)+5e^x domain of x (-infinity, infinity)

[Adi]

Hey FMHF, I'm really confused about e^x. I need to find the first and second derivative and maximum value of the function. I am told that the derivative of e^x is e^x so I found the derivatives using the method shown in the screenshot. Did I do it correctly? If yes then how do I find the maximum value of f(x)? If no then how do I do it correctly? Thanks in advance.

 

[ksdhart2]

You made a big error at this step:

with $\displaystyle a(x) = -e^{2x}, \: b(x) = 5e^{x}$

such that $\displaystyle a^\prime(x) = -e^2, \: b^\prime(x) = 5e^x$

Your derivative of b(x) is correct, but you need to double check your derivative of a(x). Being that derivatives and integrals are "opposite" functions and one "undoes" the other, we can see that your answer is not correct by integrating:

$\displaystyle \int a^\prime(x) \: dx = \int -e^2 \: dx = -e^2 \cdot x + C$

Even if we let the constant of integration be 0, this definitely does not match a(x)! Rather, you should go back and try the derivative again. Use the chain rule - let $u = 2x$ and continue...

 

[MarkFL]

A more general rule (involving the rule for exponential functions and the Chain Rule) is if we are given:

[MATH]f(x)=e^{g(x)}\implies f'(x)=e^{g(x)}g'(x)[/MATH]
And so given:

[MATH]f(x)=-e^{2x}+5e^x[/MATH]
We find:

[MATH]f'(x)=5e^x-2e^{2x}[/MATH]
Let's take a look at a plot of $f(x)$:



We see that it appears $f(x)$ has one local extremum, and it is a maximum. It occurs for $x\approx0.916$. To find the exact value, we need to solve:

[MATH]f'(x)=0[/MATH]
[MATH]5e^x-2e^{2x}=0[/MATH]
What do you get?

 

[Adi]

ok so f''(x) = 5e^x − 4e^2x?

 

[MarkFL]

ok so f''(x) = 5e^x − 4e^2x?

Yes, but I am assuming you are to use the second derivative to show that the critical value you get from the first derivative is at a maximum. So, you need to solve the equation I posted above.

 

[Steven G]

Hey FMHF, I'm really confused about e^x. I need to find the first and second derivative and maximum value of the function. I am told that the derivative of e^x is e^x so I found the derivatives using the method shown in the screenshot. Did I do it correctly? If yes then how do I find the maximum value of f(x)? If no then how do I do it correctly? Thanks in advance.

e² is a constant so its derivative is 0.

The rule you need is d(e^u)/dx = e^uu'

So for example the derivative wrt x of e^2x+3 = e^2x+3(2x+3)' = e^2x+3*2= 2e^2x+3