What is the derivative of #ln(sec^2 (x))#?
↳Redirected from
"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Mar 10, 2018
The derivative is
Explanation:
We can rewrite using trigonometry and the logarithm laws:
#y= ln(1/cos^2x) = ln(1) - ln(cos^2x) = 0 - ln((cosx)^2) = -2ln(cosx)#
Therefore by the chain rule
#y' = -2(-sinx)/cosx = (2sinx)/cosx = 2tanx#
Hopefully this helps!
