# Chain rule for variable exponents

• B
• kolleamm
In summary: Use the chain rule form of the derivative of an exponential function, which @Drakkith wrote (and I modified slightly):##\frac d{dx}\left(e^{-3x^2}\right) = e^{-3x^2} \frac d{dx}(-3x^2) = e^{-3x^2} \cdot -6x = -6xe^{-3x^2}##Thank you everyone for your help!
kolleamm
I understand that when you use the chain rule you multiply the exponent by the number in front and then reduce the power by 1. So the derivative of 2x^3 = 6x^2
I'm confused now however on how you would solve something like e^-3x, the answer turns out to be -3e^-3x

Am I missing a rule? Why isn't it -3xe^(-3x-1) ?

Thanks in advance

As you've seen, the power rule for differentiation is: ##\frac{d}{dx}x^a = ax^{a-1}##
But that's only the case when the variable is raised to an exponent that's a constant. In ##e^x## the exponent itself is the variable, making it an exponential function.
The basic rule for exponential functions is: ##\frac{d}{du}e^u = e^u\frac{d}{du}u##

For your example:
Substituting ##u## for ##-3x##, we get ##\frac{d}{du}e^u = e^u\frac{d}{du}u##
But ##\frac{d}{du}u## is ##\frac{d}{dx}-3x## which becomes ##-3##
So the original equation is: ##\frac{d}{dx}e^{-3x} = -3e^{-3x}##

jedishrfu
kolleamm said:
I understand that when you use the chain rule you multiply the exponent by the number in front and then reduce the power by 1. So the derivative of 2x^3 = 6x^2
I'm confused now however on how you would solve something like e^-3x, the answer turns out to be -3e^-3x

Am I missing a rule? Why isn't it -3xe^(-3x-1) ?
You are using a rule (the power rule) where it isn't applicable.
Power rule: ##\frac d {dx} x^n = nx^{n - 1}##
In a power function, the variable is in the base. The exponent is a constant (or at least is treated as a constant as far as the differentiation is concerned.)

##e^{-3x}## is an exponential function, not a power function. Here the base is a constant, and the exponents is a variable or a variable expression.
Drakkith said:
The basic rule for exponential functions is: ##\frac{d}{du}e^u = e^u\frac{d}{du}u##
Not quite.
##\frac{d}{dx}e^u = e^u\frac{du}{dx}##
Drakkith said:
For your example:
Substituting ##u## for ##-3x##, we get ##\frac{d}{du}e^u = e^u\frac{d}{du}u##
Correcting the above, we have ##\frac{d}{dx}e^u = e^u\frac{du}{dx} = e^{-3x} \cdot -3 = -3e^{-3x}##
Drakkith said:
But ##\frac{d}{du}u## is ##\frac{d}{dx}-3x## which becomes ##-3##
So the original equation is: ##\frac{d}{dx}e^{-3x} = -3e^{-3x}##

kolleamm and Drakkith
Mark44 said:
You are using a rule (the power rule) where it isn't applicable.
Power rule: ##\frac d {dx} x^n = nx^{n - 1}##
In a power function, the variable is in the base. The exponent is a constant (or at least is treated as a constant as far as the differentiation is concerned.)

##e^{-3x}## is an exponential function, not a power function. Here the base is a constant, and the exponents is a variable or a variable expression.
Not quite.
##\frac{d}{dx}e^u = e^u\frac{du}{dx}##
Correcting the above, we have ##\frac{d}{dx}e^u = e^u\frac{du}{dx} = e^{-3x} \cdot -3 = -3e^{-3x}##
Did you use the power rule to calculate the derivative of the exponent?

kolleamm said:
Did you use the power rule to calculate the derivative of the exponent?
No, I used the constant multiple rule. That is, ##\frac d{dx} kx = k##, so ##\frac d{dx}(-3x) = -3##.

kolleamm
Mark44 said:
No, I used the constant multiple rule. That is, ##\frac d{dx} kx = k##, so ##\frac d{dx}(-3x) = -3##.
Ah okay , what if the exponent had an exponent? For example if it was -3x^2

kolleamm said:
Ah okay , what if the exponent had an exponent? For example if it was -3x^2
Use the chain rule form of the derivative of an exponential function, which @Drakkith wrote (and I modified slightly):
##\frac d{dx}\left(e^{-3x^2}\right) = e^{-3x^2} \frac d{dx}(-3x^2) = e^{-3x^2} \cdot -6x = -6xe^{-3x^2}##

kolleamm
Thank you everyone

## What is the chain rule for variable exponents?

The chain rule for variable exponents is a mathematical rule used to find the derivative of a function that contains both a variable base and a variable exponent. It is an extension of the basic chain rule in calculus.

## How is the chain rule for variable exponents applied?

The chain rule for variable exponents is applied by first finding the derivative of the outer function (the base) and then multiplying it by the derivative of the inner function (the exponent) multiplied by the natural logarithm of the base.

## Why is the chain rule for variable exponents important?

The chain rule for variable exponents is important because it allows us to find the rate of change of a function that contains both a variable base and a variable exponent. This is especially useful in applications such as physics and economics.

## What are some common mistakes made when using the chain rule for variable exponents?

Some common mistakes made when using the chain rule for variable exponents include forgetting to include the natural logarithm of the base, mixing up the order of the functions, and not properly simplifying the expression before taking the derivative.

## Can the chain rule for variable exponents be applied to all functions?

No, the chain rule for variable exponents can only be applied to functions that contain both a variable base and a variable exponent. It cannot be applied to functions that have a constant base or exponent.

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