How do I differentiate the function F(x)=4^{3x}+e^{2x}?

[Specter]

Homework Statement

Differentiate

$F(x)=4^{3x}+e^{2x}$

Homework Equations

The Attempt at a Solution

I have an exam coming up and need some help with this problem.

$F(x)=4^{3x}+e^{2x}$

$f(x)=4^{3x}$

$G(x)=e^{2x}$

First I need to find f'(x) and g'(x), which I thought I did correctly until I found a different answer online.

Following the chain rule for exponential functions in the form $f(x)=a^{g(x)}$

$f'(x)=a^x(\ln a)(g'(x))$
$f'(x)=4^{3x}(\ln4)(3)$

Following the chain rule for functions in the form of $f(x)=e^x$

$G'(x)=e^x(\ln e)(g'(x))$
$G'(x)=e^{2x}(2)$

Sum rule:

F'(x)=f'(x)+g'(x)

$=4^{3x}(\ln 4)(3)+(2)e^{2x}$

I think I've messed up somewhere. I'm lost at this point.

 

[fresh_42]

I think I've messed up somewhere. I'm lost at this point.

Why do you think so? Looks o.k.

 

[Specter]

Why do you think so? Looks o.k.

When I type it into symbolab to check if my answer is correct, the answer they have is $f'(x)=3\ln (2) \cdot 2^{6x+1}+e^{2x}\cdot 2$

I followed all of the steps in my notes so I am not sure if I did something wrong or if the answer on that website is wrong.

 

[fresh_42]

The second term is the same, so let's have a look on the first.
You have calculated $4^{3x}(\ln 4)(3)$ which is $4^{3x}\ln (4^3)=2^{6x} \ln (2^6) = 6 \cdot 2^{6x}\ln 2 = 3\cdot 2^{1+6x}\cdot \ln 2$ and there are many more ways to write it. The symbolab solution is just the result broken down to primes. They actually should have written $3\cdot 2^{2\cdot 3 \cdot x+1}\cdot \ln 2\,$

 

[PeroK]

Homework Statement

Differentiate

$F(x)=4^{3x}+e^{2x}$

Homework Equations

The Attempt at a Solution

I have an exam coming up and need some help with this problem.

$F(x)=4^{3x}+e^{2x}$

$f(x)=4^{3x}$

$G(x)=e^{2x}$

First I need to find f'(x) and g'(x), which I thought I did correctly until I found a different answer online.

Following the chain rule for exponential functions in the form $f(x)=a^{g(x)}$

$f'(x)=a^x(\ln a)(g'(x))$
$f'(x)=4^{3x}(\ln4)(3)$

Following the chain rule for functions in the form of $f(x)=e^x$

$G'(x)=e^x(\ln e)(g'(x))$
$G'(x)=e^{2x}(2)$

Sum rule:

F'(x)=f'(x)+g'(x)

$=4^{3x}(\ln 4)(3)+(2)e^{2x}$

I think I've messed up somewhere. I'm lost at this point.

What makes you think you've messed up.

 

[PeroK]

When I type it into symbolab to check if my answer is correct, the answer they have is $f'(x)=3\ln (2) \cdot 2^{6x+1}+e^{2x}\cdot 2$

I followed all of the steps in my notes so I am not sure if I did something wrong or if the answer on that website is wrong.

In these days of spreadsheets and computers, you could always estimate the derivative and see whether your formula works.

$f'(x) \approx \frac{f(x+h) - f(x)}{h}$

For some suitably small $h$.