whats the derivative of ( x^2 +18)^2/3?

Question

please tell me the answer and the steps to get it to it

Answer 1

You can use chain rule
let t= (x^2+18)
y = (x^2+18)^2/3 = t^2/3
dy/dx = dy/dt *dt/dx
= 2/3t^(-1/3) d/dx(x^2+18)
= 2/3(x^2+18)^(-1/3) 2x
= 4x/(3(x^+18)^1/3)

Answer 2

As usual, when you take a derivative of something like x^n, the result is n(x^(n-1)). So here,
(2/3)*(x^2 + 18)^(2/3 - 1) = (2/3)*(x^2 + 18)^(-1/3)

However, what we have inside is a function of x, so we have to take the derivative of that too, and multply that by what we got before. The derivative of x^2 + 18 is 2x + 0, so this gives us:
(2/3)[(x^2 + 18)^(-1/3) ] * (2x)

Simplifying, we get the answer
(4x/3)[(x^2 + 18)^(-1/3) ]

I see some of the others keep forgetting that it's a -1/3 exponent and not a 1/3. You could alternatively write this as
(4x) / 3[(x^2 + 18)^(1/3) ]

Answer 3

2/3[(xx + 18)^(-1/3)]2x

The power rule gives you the 2/3 prefactor, and it reduces the original exponent by 3/3.

By the chain rule the derivative of (x^2 + 18) must be computed (again using the power rule, giving 2x^1); the chain rule has us multiply the two derivatives together.

Answer 4

Let y = ( x^2 +18)^2/3

By substitution,
Let u = (x^2 +18),
=> du/dx = 2x^(2-1) = 2x

=> y = u^2/3,
=> dy/du = (2/3)u^(2/3 - 1) = (2/3)u^(-1/3)

By Chain Rule,
dy/dx = dy/du * du/dx
= (2/3)u^(-1/3) * 2x
= (2/3)[(x^2 +18)^(-1/3)] * 2x
= (2/3)(2)(x)[(x^2 +18)^(-1/3)]
= (4/3)x[(x^2 +18)^(-1/3)]
= 4x / 3[(x^2 +18)^(1/3)]
or
= 4x / 3*Cuberoot(x^2 +18)

Answer 5

(2/3) (x^2 + 18) ^ (-1/3) * (2x)

= -4 x / [ 9*cuberoot (x^2 + 18)]

NOTE: The extra (2x) comes from the derivative of the inside function, chain rule)

:)

Answer 6

(2/3(x^2 +18)^-1/3)2x
= 4x/[3(x^2+18)^1/3]

Answer 7

Let y = u^(2/3) where u = x² + 18
dy/du = (2/3).u^(- 1/3)
du/dx = 2x
dy/dx = dy/du . du/dx
dy/dx = (2/3).(x² + 18)(- 1/3).2x
dy/dx = 4x / 3(x² + 18)^(1/3) = f `(x)