What is the differential of (tanx) . Can you prove it please?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We can solve the differential of tan x using the first principle, but an easier way to do it is to use the quotient rule as we know that tan x = sin x / cos x.

[tan x]' = [sin x / cos x]'

=> [(sin x)'*cos x - (sin x)*(cos x)']/(cos x)^2

the derivative of sin x = cos x and that of cos x is - sin x

=> [cos x* cos x  + sin x * sin x]/(cos x)^2

=> [(cos x)^2 + (sin x)^2]/(cos x)^2

=> 1 / (cos x)^2

=> (sec x)^2

The derivative of tan x is (sec x)^2

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Let f(x) = tanx

We need to find f'(x).

First we know that f(x) = sinx/cosx

==> We will use the qoutient rule to find the derivative.

==> f(x)= u/v such that:

u= sinx ==> u' = cosx

v= cosx ==> v' = -sinx

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

==> f'(x)= ( cosx*cosx + sinx*sinx)/ cos^2 x

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

But, we know that sin^2 x + cos^2 x = 1

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

But secx = 1/cosx.

==> f'(x) = sec^2 x

Then the derivative if tanx is (tanx)' = sec^2 x.

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