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

3 Answers

dy/dx=sec^2xdydx=sec2x

Explanation:

y=tan(x)y=tan(x)

tanx=sinx/cosxtanx=sinxcosx
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.