How do you find the derivative of #f(x) = tan^2(x)#?

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Answer 1 · 2016-07-14T02:14:23

#f(x) = sin^2x/cos^2x#

Let #f(x) = g(x)/(h(x))#, so that #g(x) = sin^2x# and #h(x) = cos^2x#. Then #f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#.

Let's differentiate #g(x)# and #h(x)#.

#g(x) = (sinx)(sinx)#

#g'(x) = cosxsinx + cosxsinx#

#g'(x) = 2cosxsinx#

#g'(x) = sin2x#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#h'(x) = (cosx)(cosx)#

#h'(x) = -sinxcosx - sinxcosx#

#h'(x) = -2sinxcosx#

#h'(x) = -(2sinxcosx)#

#h'(x) = -sin2x#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#.

#f'(x) = ((sin2x xx cos^2x) - (-sin2x xx sin^2x))/(cos^2x)^2#

#f'(x) = ((2sinxcosx xx cos^2x) - (-2sinxcosx xx sin^2x))/(cos^2x)^2#

#f'(x) = ((2sinxcos^3x) - (-2sin^3xcosx))/(cos^4x)#

#f'(x) = (2sinxcos^3x + 2sin^3xcosx)/(cos^4x)#

#f'(x) = (2sinxcosx(cos^2x + sin^2x))/cos^4x#

#f'(x) = (2sinx)/cos^3x#

#f'(x) = 2cotxsec^2x#

Hopefully this helps!