If #f(x) = ln|secx + tanx|#, what is #f'(x)#?

1 Answer
Feb 19, 2017

#d/dxln|secx + tanx| = secx#

Explanation:

Start by rewriting #secx# and #tanx# in terms of sine and cosine. Call #ln|secx + tanx|, #f(x)#.

#f(x) = ln|secx + tanx|#

#f(x) = ln|1/cosx + sinx/cosx|#

#f(x) = ln|(1 + sinx)/cosx|#

Now use the rule #ln(a/b) = lna - lnb# to simplify further.

#f(x) = ln(1 + sinx) - ln(cosx)#

You can differentiate this using the chain rule on each term. I'll show you how to do one. Let #y = lnu# and #u = 1+ sinx#. Then #dy/(du) = 1/u# and #(du)/dx = cosx#. If #dy/dx = dy/(du) * (du)/dx#, then #dy/dx = 1/u * cosx = cosx/(1 + sinx)#.

#f'(x) = cosx/(1 + sinx) - (-sinx/cosx)#

#f'(x) = cosx/(1 + sinx) + sinx/cosx#

This can be simplified.

#f'(x) = (cosx(cosx))/(cosx(1 + sinx)) + (sinx(1 + sinx))/(cosx(1 +sinx))#

#f'(x) = (cos^2x + sinx + sin^2x)/(cosx(1 + sinx))#

Apply #sin^2x + cos^2x = 1#.

#f'(x) = (1 + sinx)/(cosx(1 + sinx))#

#f'(x) = 1/cosx#

Apply #secx = 1/cosx#.

#f'(x) = secx#

Hopefully this helps!