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

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

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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##

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