How do you find the derivative of #y=1+tan^2(5x)#?

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Answer 1 · 2017-05-31T14:38:40

#y'=10sin(5x)/cos^3(5x)#

Since the derivative of a constant function is null,

and the derivative of #tan(color(blue)f(x))# is

#1/cos^2color(blue)(f(x))*f'(x)#

and the derivative of #(color(red)(g(x)^2))# is #color(red)(2g(x))*g'(x)#,

then

#y'=0+color(red)(2tan(5x))*1/cos^2(color(blue)(5x))*5#

#=10sin(5x)/cos(5x)*1/cos^2(5x)#

#=10sin(5x)/cos^3(5x)#