"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.
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
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.
Thank you both!
To differentiate an expression like x^(x^3), you can use logarithmic differentiation. Here's a step-by-step approach to find the derivative:
Step 1: Take the natural logarithm (ln) of both sides of the equation to simplify the exponent:
ln(y) = ln(x^(x^3))
Step 2: Use the logarithmic property, ln(a^b) = b * ln(a), to simplify the expression:
ln(y) = (x^3) * ln(x)
Step 3: Differentiate both sides of the equation with respect to x:
(d/dx) ln(y) = (d/dx) [(x^3) * ln(x)]
Step 4: Apply the chain rule on the right side of the equation. Let's break it down into two parts: (x^3) as the first function and ln(x) as the second function.
For the first part, differentiate (x^3) with respect to x:
d/dx (x^3) = 3x^2
For the second part, differentiate ln(x) with respect to x:
d/dx ln(x) = 1/x
Step 5: Use the chain rule to combine both parts:
(d/dx) [(x^3) * ln(x)] = [(x^3) * (d/dx) ln(x)] + [ln(x) * (d/dx) (x^3)]
Step 6: Substitute the differentiation results from Step 4 into Step 5:
(d/dx) ln(y) = (x^3) * (1/x) + ln(x) * (3x^2)
Step 7: Simplify the equation:
(d/dx) ln(y) = x^2 + 3x^2 * ln(x)
Step 8: To find the derivative of y with respect to x, multiply both sides by y and apply the property d/dx ln(y) = (1/y)(dy/dx):
(dy/dx) = y * [x^2 + 3x^2 * ln(x)]
Step 9: Substitute back y = x^(x^3):
(dy/dx) = x^(x^3) * [x^2 + 3x^2 * ln(x)]
So, the derivative of x^(x^3) with respect to x is: (dy/dx) = x^(x^3) * [x^2 + 3x^2 * ln(x)]