Question

What is the derivative of `-cosx*ln(secx+tanx)`?

Answer 1

Just do the product rule, followed by the chain rule (on only the `ln(u)`). Recall that `d/(dx)[cosx] = -sinx`, and that `d/(dx)[ln(u(x))] = 1/(u(x))(du(x))/(dx) = 1/(u(x))((d)/(dx)[u(x)])`

So:
`d/(dx)[-cosx*ln(secx+tanx)]`

`= [-cosx*(1/(secx+tanx))*(secxtanx+sec^2x)]+[ln(secx+tanx)sinx]`

Move the cosine in:
`= [(-cosx/(secx+tanx))*(secxtanx+sec^2x)]+[ln(secx+tanx)sinx]`

Factor and cancel:
`= [(-cosx/cancel(secx+tanx))*{secx(cancel(tanx+secx))}]+[ln(secx+tanx)sinx]`

Notice that `secx = 1/cosx`:
`= [-cosx*secx]+[ln(secx+tanx)sinx]`

And you get:
`= -1+sinxln(secx+tanx)`