I need help finding the derivative of

(x^2+sinx)secx

let y = (x^2 + sinx)/cosx

by quotient rule....
dy/dx = ( cosx(2x + cosx) - sinx(x^2 + cosx) )/cos^2 x

simplify if needed.

there is no quotient it is multiplication

I changed secx to 1/cosx

however, if you insist on using the product rule

dy/dx = (x^2 + sinx)(tan^2 x) + (2x + cosx)(secx)

illustrating that there is often more than one way to express the same result.

To find the derivative of the function f(x) = (x^2 + sin(x))sec(x), you can use the product rule and the chain rule.

The product rule states that if you have two functions u(x) and v(x),
the derivative of their product u(x)v(x) is given by:
(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)

First, let's differentiate the function u(x) = x^2 + sin(x):
u'(x) = d/dx ( x^2 + sin(x) )
Using the power rule and the derivative of sin(x), we get:
u'(x) = 2x + cos(x)

Next, let's differentiate the function v(x) = sec(x):
v'(x) = d/dx (sec(x))
Using the chain rule, the derivative of sec(x) is:
v'(x) = sec(x)tan(x)

Now we can use the product rule to find the derivative of f(x):
f'(x) = u'(x)v(x) + u(x)v'(x)
Substituting the values we found earlier:
f'(x) = (2x + cos(x)) * sec(x) + (x^2 + sin(x)) * sec(x)tan(x)

Simplifying further:
f'(x) = (2x + cos(x))sec(x) + (x^2 + sin(x))sec(x)tan(x)

So, the derivative of f(x) = (x^2 + sin(x))sec(x) is:
f'(x) = (2x + cos(x))sec(x) + (x^2 + sin(x))sec(x)tan(x)