Question

Last review problem for midterm.
derivative of:
(x^2cos(x^3-3x))^(1/4)

Answer 1

is cos(x^3-3x) a product term of a power

Answer 2

or*

Answer 3

the whole thing has a square root of 4 or a power of 1/4

Answer 4

\[\sqrt[4]{x ^{2}(\cos(x ^{3}-3x))}\]

Answer 5

right?

Answer 6

that is correct

Answer 7

apply the chain rule

Answer 8

consider the expression inside the square root as 'u' and differentiate first

Answer 9

any luck ?

Answer 10

Not much so do i start with
1/4(x^2cos(x^3-3x))^(-3/4) for taking the derivative of f(x) first in the chain rule

Answer 11

Then just take that times the derivative of (x^2cos(x^3-3x)) right?

Answer 12

yes... you're right . Now differentiate x^2[cos(x^3-3x)] by using the product rule

Answer 13

so 2x(cos(x^3-3x)+-sin(x^3-3x)((x^3)-3x)x^2

Answer 14

for the second term , you will have to differentiate x^3-3x as well {chain rule applies there as well } .

Answer 15

you've just taken it as it is without differentiating :)

Answer 16

it will be x^2 [ -sin(x^3-3x){d/dx(x^3-3x)}]