Hello, UMStudent!

I bet you didn't simplify the first derivative . . .If f(x) =9·ln|sec(x) + tan(x)|, . find f''(x).

. . . . . . . . sec(x)tan(x) + sec²(x) . . . . . sec(x)[tan(x) + sec(x)]

f'(x) .= .9 · -------------------------- .= .9 · --------------------------- .= .9·sec(x)

. . . . . . . . . . sec(x) + tan(x) . . . . . . . . . . sec(x) + tan(x)

Then: .f''(x) .= .sec(x)tan(x)

It's a matter of taking baby-steps . . .Suppose: F(x) = f(f(x)) .and .G(x) = [f(x)]²

Also suppose: f(4) = 5, f(5) = 2, f'(5) = 9, f'(4) = 10.

Find F'(4) and G'(4).

We have: .F(x) .= .f(f(x))

. . Then: .F'(x) .= .f'(f(x))·f'(x) . . . Chain Rule

Hence: .F'(4) .= .f'(f(4))·f'(4)

. . . . . . . . . . . . . . ↓ . . .↓

. . . . . . . . . .= . .f'(5) · 10

. . . . . . . . . . . . . .↓

. . . . . . . . . .= . . .9 · 10 . = . 90

We have: .G(x) .= .[f(x)]²

. . Then: .G'(x) .= .2·f(x)·f'(x)

Hence: .G'(4) .= .2·f(4)·f'(4)

. . . . . . . . . . . . . . .↓ . .↓

. . . . . . . . . .= . 2 · 5 · 10 . = . 100