# Prove that (tan x)' = sec^2 x

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### 3 Answers

(tan x)' = sec^2 x

Let f(x) = tanx

We know that tanx = sinx/ cosx

==> f(x) = sinx/ cosx

Now to differenctiate f(x) we will use the chain rule:

f(x) = u/v such that:

u= sinx ==> u' = cosx

v= cosx ==> v' = -sinx

==< f'(x) = (u'v- uv')/ v^2

= (cosx*cosx) - sinx*-sinx)/cos^2 x

= (cos^2 x + sin^2 x) / cos^2 x

We know that : sin^2 x + cos^2 x = 1

==> f'(x) = 1/(cos^2 x)

Also, we know that 1/cosx = sec x

==> f'(x) = (1/cosx)^2

= (sec x)^2

**Then we conclude that:**

**(tanx)' = sec^2 x **

Remember that tan(x) = sin x / cos x

So to solve the derivative of tan x, we can use the quotient rule on the right hand side of the above equation.

The quotient rule is h = f / g, then h' = f'g - fg' / g^2

D(tan x) = D( sin x / cos x ) = D(sin x ) cos x - sin x * D(cos x) / (cos x) ^2

= cos x * cos x - -sin x * sin x / (cos x)^2

= (cosx^2 + sinx^2) / (cos x ) ^2

= 1 / cos(x)^2 = sec x ^2

To find (tanx)'.

We know that (tanx)' = Lt h --> 0 {tan(x+h) - tanx}/h.

Lt h --> 0 {tan(x+h) - tanx}/h = Lth-->0{[(tanx+tanh)/(1-tanxtanh)-tanx]/h

Lt h --> 0 {tan(x+h) - tanx}/h = Lth-->0 { (tanx+tanh - tanx + tan^2xtanh}/h(1+tanxtanh)

Lt h --> 0 {tan(x+h) - tanx}/h = Lth--> 0 (tanh+tan^2xtanh)/ h(1+tanxtanh)

Lt h --> 0 {tan(x+h) - tanx}/h = Lth --> 0 (tanh/h) (1+tan^2x)/(1+tanxtanh.

= Lt h--> 0 (sinx/h)(1/cosh) sec^2x/(1+tanxtanh).

Lt h --> 0 {tan(x+h) - tanx}/h = ( 1/cos0)(sec^2x)/(1-tanx *0) , as Lt--> 0(sinh/h) = 1.

Lt h --> 0 {tan(x+h) - tanx}/h = sec^2x.

Therefore (tanx)' = (secx)^2.