Question

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

Answer 1

`y=tanx\\\\y'=\lim\limits_(h\to0)(tan(x+h)-tanx)/(h)=\lim\limits_(h\to0)(1)/(h)\left((sin(x+h))/(cos(x+h))-(sinx)/(cosx)\right)\\\\=\lim\limits_(h\to0)(1)/(h)\left((sin(x+h)cosx-sinxcos(x+h))/(cosxcos(x+h))\right)=\lim\limits_(h\to0)(sin(x+h-x))/(hcosxcos(x+h))\\\\=\lim\limits_(h\to0)(sinh)/(hcosxcos(x+h))=\left[(0)/(0)\right]\xrightarrow{l'Hopital's\ rule}\lim\limits_(h\to0)(cosh)/(cosxcos(x+h)+hcosx[-sin(x+h)])\\\\=(cos0)/(cosxcos(x+0))=(1)/(cos^2x)`