How do you find the derivative of ##y=tan(x)## using first principles?

There are two possible answers depending upon what you mean by "first principles."

1) If by first principles, you mean using derivatives we have already found (like the derivative of y = sin x and y = cos x) and rules we already know (like the quotient rule) this is relatively straightforward.

We know from trigonometry that ##y = tan(x) = (sin(x)) / (cos(x))##

Now we find the derivative of this expression using the :

##y' = ((cos(x) * cos(x) - sin(x) * (-sin(x)))/(cos^2(x)))##

##y'=(cos^2(x)+sin^2(x))/(cos^2(x))##

##y' = 1/(cos^2(x)) = sec^2(x)##

2) If by first principles, you mean that you would like to go back to the definition of the derivative, then we have to do a bit more work.

##y' = lim_(h rarr0)((tan(x+h)-tanx)/h)##

##y' = lim_(h rarr0)((((tanx+tan h)/(1-tanx*tan h))-tanx)/h)## using the identity for tan (a + b) from trigonometry

##= lim_(h rarr0)(((tanx+tan h-tanx+tan^2(x)tan h)/(1-tanxtan h))/h)##

##= lim_(h rarr0)((tan h+tan^2xtan h)/(h*(1-tanxtan h)))##

##= lim_(h rarr0)(1+tan^2x)/(1-tanxtan h)*cancel(lim_(h rarr0) tan h/h)^(1)##

Note that ##lim_(hrarr0)(tan h/h)=1## because

##lim_(hrarr0)(tan h/h)##=##lim_(hrarr0)(sin h/(cos h*h))##=

##lim_(hrarr0)(sin h/h)*lim_(hrarr0)(1/cos h)##

=1*1=1 (the first limit is a famous one, proven by the squeeze theorem) that you probably learned in your calculus class, while the second limit can be found by simply substituting zero for h)

= ##lim_(hrarr0) (1+tan^2x)/(1-tanx*0)##

##= 1+tan^2x= sec^2x##