Question

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

Answer 1

`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`

Answer 2

`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`

Answer 3

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`.