Find y' for y=7sec^3x

Taking the derivative of 7 times secant of x^3:

We take out 7 as a constant focus on secant (x^3)

To take the derivative, we use the chain rule, taking the derivative of the inside, bringing it out, and then the derivative of the original function. For example:

The derivative of x^3 is 3x^2, and the derivative of secant is tan (x) and sec (x).

Knowing this: secant (x^3) becomes tan (x^3) * sec (x^3) * 3x^2. We transform tan (x^3) into sin (x^3) / cos (x^3) since tan (x) = sin (x) / cos (x). Then secant (x^3) becomes 1/cos (x^3) since the secant is the reciprocal of the cosine.

We then multiply everything together to simplify:

sin (x^3) * 3x^2 / cos (x^3) * cos (x^3) becomes

3x^2 * sin (x^3) / (cos (x^3)) ^2

and multiplying the constant 7 from the beginning:

7 * 3x^2 = 21x^2, so ...

our derivative is 21x^2 * sin (x^3) / (cos (x^3)) ^2