Question

Explain the two types of the chain rule of calculus when finding derivatives. How do i identify the inside/outside function and how do i determine when to use the different types?

Answer 1

Please and thank you :)

Answer 2

For example, applying the chain rule to \[e^{-2t}\] results in \[-2e ^{-2t}\]. what type of chain rule is this using?

Answer 3

pleeease. medals, think of the medal :D

Answer 4

For your example, think of it this way:
\[f(t)=e^t\\
g(t)=-2t\\
\text{So, }f(g(t))=f(-2t)=e^{-2t}\]
By the chain rule, the derivative of \(f(g(t))=f'(g(t))\cdot g'(t)\). So, you have
\[f'(g(t))=f(g(t)),\text{ since }f'(t)=f(t)=e^t,\\
g'(t)=-2,\text{ and so,}\\
\left(f(g(t))\right)'=f(g(t))\cdot g'(t)=e^{-2t}\cdot(-2)=-2e^{-2t}\]

Answer 5

so for this example why is the exponent the inside function?

Answer 6

There's another way of writing the exponential function that will help to see why. In some textbooks, you'll see the function \(e^{h(x)}\) written as \(\exp(h(x))\).

For your function, you'd assign \(f(t)=\exp(t)\) and \(g(t)=-2t\), which gives you \(f(g(t))=\exp(-2t)=e^{-2t}.\) It should be clearer that -2t is the inside function.

Answer 7

ohhh i see. can i apply this to other equations to tell if the exponent is inside or outside?

Answer 8

h(z)=2(4z+\[h(z)=2(4z+e ^{-8z})^{-10}\]
The inside function here is \[(4z+e ^{-8z}) \] or does this only apply when its e to the power of an exponent?

Answer 9

Here are some more examples:
\[\sin(2x)\Rightarrow f(x)=\sin(x),\;g(x)=2x\\
\begin{align*}(\sin(2x))'&=(\sin)'(2x)\cdot (2x)'\\
&=\cos(2x)\cdot 2\\
&=2\cos(2x)\end{align*}\]
You probably won't see the first step in a textbook. I just wrote it that way to make it a bit more obvious as to what you're doing.

Here's a more complicated trig function:
\[\sec^2(\tan x)\Rightarrow f(x)=x^2,\;g(x)=\sec x, \;h(x)=\tan x,\;j(x)=x\\
\begin{align*}\left(\sec^2(\tan x)\right)'&=\left(^2\right)'\left(\sec(\tan x)\right) \cdot \left(\sec\right)'(\tan x) \cdot \left(\tan x\right)' \cdot \left(x\right)'\\
&=2\sec(\tan x)\cdot\sec(\tan x)\tan(\tan x)\cdot\sec^2x\cdot 1\\
&=2\sec^2(\tan x)\tan(\tan x)\sec^2x\end{align*}\]
Again, the notation is proper here, so I wouldn't advise using this on a test unless it helps you understand the process.

And as a last example, in response to your most recent reply, I'll do one with a function in the exponent:
\[\large x^3e^{x^2}\Rightarrow f(x)=e^x,\;g(x)=x^2\\
\large\begin{align*}\left(x^3e^{x^2}\right)'&=\large \left(x^3\right)'e^{x^2}+x^3\left(e^{x^2}\right)'\\
&=3x^2e^{x^2}+x^3\left(\exp(x^2)\right)'\\
&=3x^2e^{x^2}+x^3\left[\left(\exp\right)'(x^2)\cdot\left(x^2\right)'\right]\\
&=3x^2e^{x^2}+x^3\left[\exp(x^2)\cdot\left(2x\right)\right]\\
&=3x^2e^{x^2}+2x^4\exp(x^2)\\
&=3x^2e^{x^2}+2x^4e^{x^2}\end{align*}\]
(Note the use the product rule, of course.)

Answer 10

oh mah gerd. thank you. I get it.

Answer 11

I just realized, color-coding the inside/outside functions would have been more of a visual aid. Let me know if you have trouble understanding what I wrote just now.

Answer 12

no i understand. you applyed it so that
\[x^{3}\exp((x ^{2}))\]
so you have x^2 as the inside.

Answer 13

^.^ thank you so much.

Answer 14

You're welcome!

Answer 15

oh wait! one last question D: why is it that you dont find the derivative of e to be 0?

Answer 16

That would be the case if you were given \(e=e^1.\) Here, the exponent is fixed, so you're basically taking the derivative of the constant \(e\), which is just 0.

On the other hand, if you're given \(e^x\), the exponent is a variable so the same derivative rules don't apply. Now, the derivative is \((e^x)'=e^x.\)

Basically, the difference lies in the fact that one is a constant and the other is a function (i.e. dependent on a variable).

You can also think about the limit definition of the derivative. When you take a derivative, you are getting the rate of change of output as you change the input. Given a constant function (i.e., for all values of \(x\), \(f(x)=\text{constant}\)), the output stays the same regardless of the input. Meanwhile, for a non-constant function, the output will almost always changes for small changes in x.

Answer 17

How do i give you all my medals?

Answer 18

:P thank you

Answer 19

You're welcome!