Question #3b1e3

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Answer 1 · 2017-01-05T13:41:55

Given #f(x)=sec^2x=2/(2cos^2x)#

#=>f(x)=2/(1+cos2x)#

#f'(x)=stackrel(Lt)(""_(hto0)) (f(x+h)-f(x))/h#

#=2stackrel(Lt)(""_(hto0))1/h(1/(1+cos(2(x+h)))-1/(1+cos(2x)))#

#=2stackrel(Lt)(""_(hto0)) 1/h(cos(2x)-cos(2(x+h)))/((1+cos(2(x+h)))(1+cos(2x))))#

#=2 stackrel(Lt)(""_(hto0)) 1/h(cos(2x)-cos(2(x+h)))/((1+cos(2(x+h)))(1+cos(2x))))#

#=2stackrel(Lt)(""_(hto0))1/h(2sin(2x+h)sinh)/((1+cos(2(x+h)))(1+cos(2x))))#

#=(4sin(2x))/((1+cos(2x))(1+cos(2x)))#

#=(8sinxcosx)/(2cos^2x.2cos^2x)#

#=2sec^2xtanx#