How to Differentiate x^x ? [2 Different Methods]

There are 2 different ways to take the derivative of x^x, which are implicit differentiation, and the chain rule. In this video, we will be solving for the derivative of y=x^x by using these two methods. For the implicit differentiation, we first take the natural log on both sides of the equation, and we are able to apply implicit differentiation to solve for the derivative. For the chain rule, we also take the natural log on both sides, but the difference is that instead of taking the derivative on both sides directly, we first rewrite the equation from logarithmic form into exponential form, so that we can apply the chain rule.

#derivatives #differentiation #calculus #derivative

Implicit Differentiation Explanation:

Chain Rule Explanation:

TIMECODES:
0:00 Intro
0:14 First Method - Implicit Differentiation
2:17 Second Method - Chain Rule
4:12 Outro

[itsale]

You can also rewrite x^x as e^(xlnx) since this is in exponential form. From there, the term e^(xln(x)) stays the same and you multiply it by the derivative of xln(x) (use product rule). On the right side, you get e^(xlnx) * (1+lnx) and on the left side, the derivative of y with respect to x is 1 dy/dx. Rewrite e^(xlnx) as x^x and you get the same answer as here.

[adwz]

Awesome, these are two fundamental methods everyone should know about

Now differentiate x^^3 = x^(x^x) B)

[Sakibmanzoor-mu]

I didn't know we could solve it using the chain rule . Thanks, pal !

[anotherelvis]

Or with partial derivatives:
d/dx x^x = d/dy y^x |y=x + d/dy x^y |y=x
= ln(y)*y^x |y=x + y*x^(y-1) |y=x
= ln(x)*x^x + x*x^(x-1)
= (ln(x) + 1) *x^x

[josepherhardt]

I know it doesn't work, but I'd've liked to have had at least a brief discussion about why using the exponent function derivative doesn't work, e.g.,
we know y = x ^ n has derivative y' = nx^(n-1)
If we attempt this with y = x^x, we get y' = x(x^(x-1)) * dx/dx = x^(x-1+1) * 1 = x^x
I guess my question here is, _exactly what rule is it we're violating when we attempt this?_

[victor]

There is a third method. Differentiation of function of two variables.

[dacooooord]

I just automatically assumed x=exp(ln(x)) and we just use the rules of derivatives to get y'= ln(x)exp(ln(x)×x)=ln(x)x^x but you said that it was (1+ln(x))x^x i just cant figure where im wrong

[Spyro]

Why is natural log used Instead of normal log?

[gelbkehlchen]

1.Solution:
y = x^x with 0<x |ln() ⟹

ln(y) = ln(x^x) = x*ln(x) |()’ ⟹
1/y*y’ = 1*ln(x)+x*1/x = ln(x)+1 |*y ⟹
y’ = [ln(x)+1]*y = [ln(x)+1]*x^x

2.Solution:
y = x^x with 0<x |e^() ⟹

y = e^[ln(x^x)] = e^[x*ln(x)] |()’ ⟹

y’ = e^[x*ln(x)]*[1*ln(x)+x*1/x] = e^[ln(x)*x]*[ln(x)+1] ⟹
y’ = [e^ln(x)]^x*[ln(x)+1] = x^x*[ln(x)+1]

[zachansen]

sad this wasn't a fish to the fish video :(