How to Differentiate x^x | Essential Calculus #1
Vložit
- čas přidán 11. 09. 2024
- 🌟 Today, I'm breaking down the differentiation of two mind-bending functions: x^x and x^x^x. These aren't your everyday derivatives, and mastering them will give you serious bragging rights in any maths circle.
In this video, I'll guide you step-by-step through the differentiation process, making complex concepts easy to understand. By the end, you'll not only know how to differentiate x^x and x^x^x, but you'll also gain a deeper appreciation for the elegance of calculus.
Whether you're a student prepping for exams, a math enthusiast, or just curious about higher-level calculus, this video is for you. Make sure to like, subscribe, and hit the bell icon so you never miss out on more awesome math content! And if you have any questions or topics you'd like me to cover next, drop a comment below! 📘✏️
or using chain rule:
y = e^(x ln x)
y' = e^(x ln x)(ln x + 1) = x^x(ln x + 1)
and again:
y = e^(x^x ln x)
y' = e^(x^x ln x)(x^x(ln x + 1) ln x + x^(x - 1))
y' = x^(x^x)((x^x)((ln x)^2 + ln x + 1/x)) = x^(x^x + x - 1)(x (ln x)^2 + x ln x + 1)
but it's easier to use implicit differentiation as you have done
The first time I made it without any "rule", just by definition BY HAND, dumbest or form the smartest things I've ever done
Amazing!
Thanks!
3:42 I think of implicit differentiation as an application of the chain rule. Multiply ln's derivative with respect to its argument by the derivative of its argument with respect to x. d(ln y)/dy * d(y)/dx = 1/y * dy/dx.
thank you so much!!
Another way could been as follows:
Notice that x^x = e^log(x^x)= e^xlogx
The derivativa of y(x)= e^(f(x)) = f'(x)e^f(x)
Substituting we obtain
y = e()
Can you make a video on The Reimann Sphere?
This sounds interesting, I’ll look in to it
Can X^X^X × X^X = X^X^X+1 ??
It would be x^(x^(x) +1)
You should use pokemon music in background, its worked for me a lot when doing calc!