Derivatives?

Question

The derivative of sqrt of 1-3x^3 is....

How do you find the derivative of something under a square root. please do it in steps.

Thanks!!

Answer 1

the differentiation formula for a square root is (1)/(2 √u) multiplied by (du/dx)

in this case, the 'u' is '1-3x^2' note that the "u" is the variables/constant inside the square root.

dy / dx = 1 / {(2 √(1-3x^2)) x [0-3(3)(x)^(3-1))]}

by simplifying you'll have the answer,

-9x^2 over 2√(1-3x^3)

by rationalization, you'll arrive with

-9x^2 (2√(1-3x^3)) over -6(x^3) - 2

Answer 2

Actually the derivative is:
(-9/2*x^2)*(1-3*x^3)^(-1/2)

Answer 3

Assuming that :-
y = (1 - 3 x ³)^(1 / 2)
dy / dx = (1 / 2)(1 - 3 x ³)^(- 1 / 2) (- 9 x ²)
dy / dx = (- 9 / 2) x ² / √ ( 1 - 3 x ³ )

Answer 4

take the expression "1-3x^3" and raise it to the 1/2 power....

this is "U" raised to the 1/2

so .... the deriv is "dy/du ... du/dx"
or ...... (1-3x^3)^-1/2 .... -9x^2

it looks like this
(-9x^2) / (1-3x^3)^1/2