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

• Specter
In summary: 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\,## :wink:
Specter

## Homework Statement

Differentiate

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

## 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.

Specter said:
I think I've messed up somewhere. I'm lost at this point.
Why do you think so? Looks o.k.

fresh_42 said:
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.

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\,##

Specter
Specter said:

## Homework Statement

Differentiate

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

## 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.

Specter
Specter said:
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##.

Specter

## What is differentiation?

Differentiation is a mathematical process used to find the rate of change of a function with respect to its independent variable. It involves finding the derivative of a function, which represents the slope of the tangent line at a given point on the function.

## Why is differentiation important?

Differentiation is important because it allows us to analyze and understand the behavior of functions. It is used in various fields such as physics, economics, and engineering to model and solve real-world problems. It also helps in finding maximum and minimum values of a function, which is useful in optimization and decision-making.

## What are the basic rules of differentiation?

The basic rules of differentiation include the power rule, product rule, quotient rule, and chain rule. The power rule states that the derivative of a function raised to a power is equal to the power multiplied by the function raised to the power minus one. The product rule states that the derivative of a product of two functions is equal to the first function times the derivative of the second, plus the second function times the derivative of the first. The quotient rule states that the derivative of a quotient of two functions is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. The chain rule states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

## How do I differentiate a function?

To differentiate a function, you need to apply the basic rules of differentiation. Start by identifying the independent variable in the function. Then, use the appropriate rule(s) to find the derivative of the function. Remember to simplify your answer and use the chain rule if the function is a composition of two or more functions.

## What is the difference between differentiation and integration?

Differentiation and integration are inverse operations of each other. While differentiation is used to find the rate of change of a function, integration is used to find the area under the curve of a function. In other words, differentiation tells us how the function changes, while integration tells us how much the function has changed. Another key difference is that differentiation results in a function, while integration results in a number.

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