Find the derivative of
y = 3x² · sec x
To differentiate this function, apply the product rule:
y = u · v ⇒ y' = u' · v + u · v'
So, for this one, you have
y' = (3x²)' · sec x + 3x² · (sec x)'
y' = 3 · (x²)' · sec x + 3x² · (sec x)'
The derivative of x² is 2x:
y' = 3 · 2x · sec x + 3x² · (sec x)'
y' = 6x · sec x + 3x² · (sec x)'
The derivative of sec x is sec x · tan x:
y' = 6x · sec x + 3x² · (sec x · tan x)
Take out the common factor 3x · sec x:
y' = (3x · sec x) · 2 + (3x · sec x) · x · tan x
y' = (3x · sec x) · (2 + x · tan x) <——— this is the answer.
I hope this helps. =)
