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
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.
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
Amazing!
Thanks!
The first time I made it without any "rule", just by definition BY HAND, dumbest or form the smartest things I've ever done
thank you so much!!
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.
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!