d/dx u^v = v u^(v-1) du/dx + u^v ln u dv/dx
here
u = x
v = x^3
du/dx = 1
dv/dx = 3x^2
so
x^3 x^(x^3-1) + 3 x^2 x^(x^3) ln x
x^(x^6-1) + 3 x^(x^3+2)ln x
"Differentiate x^(x^3)"
I can do it when the base is a number, but I'm confused as to how to do it with x as the base.
3 answers
Take log on both sides and use implicit differentiation. Solve for y'.
y=x^(x^3)
ln(y)=(x^3)ln(x)
differentiate both sides with respect to x:
(1/y)(dy/dx) = (3x²)ln(x)+x³/x
y'/y = x²(3ln(x)+1)
y' = y x²(3ln(x)+1)
= x^(x^3) x²(3ln(x)+1)
Check my work.
y=x^(x^3)
ln(y)=(x^3)ln(x)
differentiate both sides with respect to x:
(1/y)(dy/dx) = (3x²)ln(x)+x³/x
y'/y = x²(3ln(x)+1)
y' = y x²(3ln(x)+1)
= x^(x^3) x²(3ln(x)+1)
Check my work.
Thank you both!