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How do you find the derivative of # y = ln(secx)#?

2 Answers
Apr 1, 2017

Answer:

See below

Explanation:

In order to differentiate this function, we must use the chain rule:

#(f g(x))'=f'(g(x))xxg'(x)#.

Informally, this means that if we have to derivate a composite function, #fg(x)#, then we differentiate #f# with respect to #x#, treating #g(x)# as if it were #x# and then multiply that derivative by #g'(x)#.

The function to derivate is #lnsecx#. So #(lnsecx)'=ln'(secx)xx(secx)'=1/secx xxsecxtanx=tanx#

Apr 1, 2017

Answer:

#dy/dx=tanx#

Explanation:

We can also rewrite this using logarithm rules:

#y=ln(secx)=ln(1/cosx)=ln((cosx)^-1)=-ln(cosx)#

The derivative of #ln(x)# is #1/x#, so according to the chain rule the derivative of #ln(f(x))# is #1/f(x)*f'(x)#.

Thus,

#dy/dx=-1/cosx*d/dxcosx#

The derivative of #cosx# is #-sinx#:

#dy/dx=-1/cosx(-sinx)=sinx/cosx=tanx#