High School Chain rule for variable exponents

[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

 

[Drakkith]

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: $\frac{d}{dx}e^{-3x} = -3e^{-3x}$

 

[Mark44]

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.

The basic rule for exponential functions is: $\frac{d}{du}e^u = e^u\frac{d}{du}u$

Not quite.
$\frac{d}{dx}e^u = e^u\frac{du}{dx}$

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 $\frac{d}{dx}e^u = e^u\frac{du}{dx} = e^{-3x} \cdot -3 = -3e^{-3x}$

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]

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?

 

[Mark44]

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]

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

 

[Mark44]

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