By definition If #y=f(x)# then:
# dy/dx=f'(x) =lim_(hrarr0)( (f(x+h)-f(x))/h ) #
So, with # y=tanx # we have:
# dy/dx = lim_(h rarr0)((tan(x+h)-tanx)/h) #
Using the trig identity for #tan (a + b)# we have;
# dy/dx = lim_(h rarr0)((((tanx+tan h)/(1-tanx*tan h))-tanx)/h)#
Putting over a common denominator of #1-tanxtan h#:
# :. dy/dx = lim_(h rarr0)((((tanx+tan h) - tanx(1-tanxtan h))/(1-tanxtan h))/h)#
# :. dy/dx = lim_(h rarr0)(((tanx+tan h-tanx+tan^2xtan h)/(1-tanxtan h))/h)#
# :. dy/dx = lim_(h rarr0)((tan h+tan^2xtan h)/(h(1-tanxtan h)))#
# :. dy/dx = lim_(h rarr0)((tan h(1+tan^2x))/(h(1-tanxtan h)))#
# :. dy/dx = lim_(h rarr0)((1+tan^2x)/(1-tanxtan h) * tan h/h)#
# :. dy/dx = lim_(h rarr0)((1+tan^2x)/(1-tanxtan h)) * lim_(h rarr0)(tan h/h)#
Now,
# lim_(hrarr0)(tan h/h)=lim_(hrarr0)(sin h/(cos h*h))#
# :. lim_(hrarr0)(tan h/h)=lim_(hrarr0)(sin h/h)*lim_(hrarr0)(1/cos h)#
# :. lim_(hrarr0)(tan h/h)=1*1# (standard trig limits)
# :. lim_(hrarr0)(tan h/h)=1#
And so,
# :. dy/dx= lim_(h rarr0)((1+tan^2x)/(1-tanxtan h)) #
# :. dy/dx = (1+tan^2x)/(1-0) #
# :. dy/dx = 1+tan^2x #
And finally, using the standard trig identity #1+tan^2A-=sec^2A# we get the solution;
# dy/dx = sec^2x #
