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

3 Answers

#dy/dx=sec^2x#

Explanation:

#y=tan(x)#

#tanx=sinx/cosx#
#y+Deltay=tan(x+Deltax)#
#tan(x+Deltax)=sin(x+Deltax)/cos(x+Deltax)#
#y+Deltay-y=tan(x+Deltax)-tan(x)#
#sin(A+B)=sinAcosB+cosAsinB#
#cos(A+B)=cosAcosB-sinAsinB#

#tan(x+Deltax)=(sinxcosDeltax+cosxsinDeltax)/(cosxcosDeltax-sinxsinDeltax)#

#Deltay=(sinxcosDeltax+cosxsinDeltax)/(cosxcosDeltax-sinxsinDeltax)-sinx/cosx#

#Deltay=(((cosx(sinxcosDeltax+cosxsinDeltax)-sinx(cosxcosDeltax-sinxsinDeltax)))/(cosx(cosxcosDeltax-sinxsinDeltax)))#

#=(cosxsinxcosDeltax+cos^2xsinDeltax-sinxcosxcosDeltax+sin^2xsinDeltax)/(cos^2xcosDeltax-cosxsinxsinDeltax)#

#=(cos^2xsinDeltax+sin^2xsinDeltax)/(cos^2xcosDeltax-cosxsinxsinDeltax)#

#=((cos^2x+sin^2x)sinDeltax)/(cos^2xcosDeltax-cosxsinxsinDeltax)#

Dividing throughout by
#cos^2xcosDeltax#

#=(tanDeltax+tan^2xtanDeltax)/(1-tanxtanDeltax)#

#Deltay=(1+tan^2x)(tanDeltax)/(1-tanxtanDeltax)#

#(Deltay)/(Deltax)=1/(Deltax)xx(1+tan^2x)(tanDeltax)/(1-tanxtanDeltax)#

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

#(Deltay)/(Deltax)=sec^2x xx1/(Deltax)xx(tanDeltax)/(1-tanxtanDeltax)#

applying limits as #Deltax->0#

#lim(Deltay)/(Deltax)=lim(sec^2x xx1/(Deltax)xx(tanDeltax)/(1-tanxtanDeltax))#

#=sec^2x xx lim(tanDeltax)/(Deltax)/(1-tanx xx limtanDeltax)#

#lim(tanDeltax)/(Deltax)=1#

#lim(tanDeltax)=0#

Thus,

#dy/dx=sec^2x xx 1/(1-tanx xx 0)#

#dy/dx=sec^2x#

Jul 6, 2018

#d/dx tanx = sec^2x#

Explanation:

By definition:

#d/dx tanx = lim_(h->0) (tan(x+h)-tanx)/h#

Using the trigonometric formulas for the sum of two angles:

#tan(x+h) = sin(x+h)/cos(x+h)#

#tan(x+h) = (sinxcos(h)+cosx sin(h))/(cosxcos(h)-sinxsin(h))#

#tan(x+h) = ((sinxcos(h)+cosx sin(h))/(cosxcos(h)))/((cosxcos(h)-sinxsin(h))/(cosxcos(h)))#

#tan(x+h) = (tanx+tan(h))/(1-tanx tan(h))#

So:

#d/dx tanx = lim_(h->0) ((tanx+tan(h))/(1-tanx tan(h))-tanx)/h#

#d/dx tanx = lim_(h->0) (cancel(tanx)+tan(h)-cancel(tanx)+tan^2x tan(h))/(h(1-tanx tan(h))#

#d/dx tanx = lim_(h->0) (tan(h)(1+tan^2x ))/(h(1-tanx tan(h))#

#d/dx tanx = (1+tan^2x ) lim_(h->0) tan(h)/h1/(1-tanx tan(h))#

and as:

# lim_(h->0) tan(h)/h = 1#

#lim_(h->0) 1/(1-tanx tan(h)) = 1#

#d/dx tanx = 1+tan^2x #

Or equivalently:

#d/dx tanx = 1+sin^2x /cos^2x = (cos^2x+sin^2x)/cos^2x = 1/cos^2x = sec^2x #

Jul 6, 2018

Please go through The Explanation.

Explanation:

By Definition, # d/dx[f(x)]=lim_(t to x) {f(t)-f(x)}/(t-x)#.

#:. d/dx[tanx]=lim_(t to x) (tant-tanx)/(t-x)#,

#=lim{sint/cost-sinx/cosx}/(t-x)#,

#=lim(sintcosx-costsinx)/{(t-x)costcosx}#,

#=limsin(t-x)/(t-x)*1/cost*1/cosx#.

Since, #lim_(theta to 0) sintheta/theta=1#, and, #cos# function is continuous, we have,

#d/dx[tanx]=1*1/cosx*1/cosx=1/cos^2x=sec^2x#.