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

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There are two approvely answers depending upon what you moderation by "earliest principles."

1) If by earliest principles, you moderation using derivatives we own already set-up (approve the derivative of y = sin x and y = cos x) and administrations we already perceive (approve the quotient administration) this is relatively sincere.

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

Now we ascertain the derivative of this countenance 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 earliest principles, you moderation that you would approve to go tail to the determination of the derivative, then we own to do a bit further product.

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

##y' = lim_(h rarr0)((((tanx+tan h)/(1-tanx*tan h))-tanx)/h)## using the oneness 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## accordingly

##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 earliest word is a illustrious one, proven by the press theorem) that you probably read in your calculus rank, while the promote word can be set-up by merely substituting nothing for h)

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

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

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