Question

How do you take the derivative of this?

Answer 1

\[e^({x})^2\]

Answer 2

\(\color{#000000}{ \displaystyle e^{x^2} }\) ?

Answer 3

e raised to the x raised to the 2nd power...idk how to do that on here

Answer 4

yes

Answer 5

You just need two know 2 things.

1) The derivative of e^x is e^x.
2) The Chain Rule

Answer 6

Have you done the chain rule previously?

Answer 7

oh ok so bring the 2 to the front

Answer 8

\(\color{#000000}{ \displaystyle \frac{d}{dx} e^x=e^x }\)

Answer 9

(d/dx is the notation for the derivative, I will assume you are familiar with it, unless you tell me otherwise.)

Answer 10

So, the first line reads as
«The derivative of \(e^x\) is \(e^x\)»
(this can be verified through the first principles)

Answer 11

Now, the chain rule.

Basically, if you are differentiating a composite function \(H(f(x))\), then you first differentiate the whole function H with respect to the argument \(g(x)\), and then differentiate \(g(x)\) with respect to the argument \(x\).

In other words,
\(\color{#000000}{ \displaystyle \frac{d}{dx} H(f(x))=H'(f(x))\times g'(x) }\)

you can prove this with a delta-epsilon proof,
you can reason intuitively, (dy/dx=dy/du • du/dx),
but I won't go into proving it now...
Rather, I will give you an example.

Answer 12

ohhhhhh

Answer 13

so like the chain rule.... 2 * e^x * e ^x

Answer 14

because the inside is e^x and you have to take the derivative of that which like you said is e^x

Answer 15

or no

Answer 16

Suppose that I want to take the derivative of
\(\color{#000000}{ f(x)=\sqrt{x^4+3} }\).

My other friends like the product and quotient
are obviously helpless here. We will use the chain
rule!!

[1] Differentiate the whole function with respect
to the inner argument (\(x^4+3\) in this case).

What is the derivative of \(\sqrt{x}\) ? It is \(\frac{1}{2\sqrt{x}}\).
So, we will say that the derivative in our case is, \(\color{#000000}{ \frac{1}{2\sqrt{x^4+3}} }\).

[2] Our function, however, was not just \(\sqrt{x}\).
We had an argument inside the root, the \(x^4+3\).

So, will have to multiply our derivative times the derivative of the inner argument.
\(\color{#000000}{ \frac{1}{2\sqrt{x^4+3}} \times \left(\frac{d}{dx}[x^4+3]\right) }\)

\(\color{#000000}{ \frac{1}{2\sqrt{x^4+3}} \times \left(4x^3+0\right) }\)

and all that is left, is to simplify,
\(\color{#000000}{ \frac{2x^3}{\sqrt{x^4+3}} . }\)

Answer 17

So, recap for this example, without lengthy wording ...


My function: \(\color{#000000}{ f(x)=\sqrt{x^4+3} }\)
I want the derivative: \(\color{#000000}{ f'(x)=~? }\)


Since, the derivative of \(\color{#000000}{ \sqrt{x} }\) is \(\color{#000000}{ \frac{1}{2\sqrt{x}} }\), \(\\[1.1em]\)
therefore, the derivative of \(\color{#000000}{ \sqrt{x^4+3} }\) is \(\color{#000000}{ \frac{1}{2\sqrt{x^4+3}} }\), \(\\[1.1em]\)
and because I have another argument inside,
I multiply times the derivative of that argument.

The total derivative of \(\color{#000000}{ \sqrt{x^4+3} }\) is:
\(\color{#000000}{ \frac{1}{2\sqrt{x^4+3}}\times \left(\frac{d}{dx}~x^4+3 \right) = }\) \(\\[1.1em]\)
\(\color{#000000}{ \frac{1}{2\sqrt{x^4+3}}\times \left(4x^3 \right) = \frac{4x^3}{2\sqrt{x^4+3}} =\frac{2x^3}{\sqrt{x^4+3}} .}\) \(\\[1.1em]\)

Answer 18

Another more similar example ....

Answer 19

The derivative of \(\color{#000000}{ \displaystyle e^{x^3} }\).

The derivative of \(e^x\) is \(e^x\), so the derivative of \(\color{#000000}{ \displaystyle e^{x^3} }\) is \(\color{#000000}{ \displaystyle e^{x^3} }\), and I will need to multiply times the derivative of the inner argument (of \(x^3\)). So, my total derivative is,

\(\color{#000000}{ = e^{x^3} \times \left(\frac{d}{dx}~x^3\right) =e^{x^3} \times \left(3x^2\right) =3x^2e^x. }\)

Answer 20

again, the bottom line,

you differentiate \(\color{#000000}{ H(f(x)) }\), and treat \(\color{#000000}{ f(x) }\) like a variable (like \(x\)),
and then you multiply times the derivative of the inside, times \(f'(x)\).

Answer 21

Thank you so much for the very detailed and thorough explanation! :D

Answer 22

No problem.