Prove that the derivative of tan x is sec^2 x.

Asked on by tonys538

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ishpiro's profile pic

ishpiro | College Teacher | (Level 1) Educator

Posted on

Alternatively, one can find the derivative of tan(x) by using the identity

`tan(x) = sin(x)/cos(x)`

To find the derivative of a fraction, apply the quotient rule:

`(f/g)' = (f'g - fg')/g^2`

Here, f would be sin(x) and g would be cos(x). The derivatives of sine and cosine are

(sin(x))' = cos(x)

(cos(x))' = -sin(x)

Substituting these functions into the quotient rule, we get

`tan(x)' = (sin(x)/cos(x))' = (cos(x)cos(x) - sin(x)(-sin(x)))/(cos(x))^2 = ((cos(x))^2 + (sin(x))^2)/(cos(x))^2`

According to the Pythagorean identity,

`(cos(x))^2 + (sin(x))^2 = 1`

So `tan(x)' = 1/(cos(x))^2`

According to the cofunction identity, `sec(x) = 1/cos(x)` , so

`tan(x)' = (sec(x))^2` .

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The derivative of a function f(x) can be derived from first principles as the limit `lim_(h->0) (f(x+h) - f(x))/h` .

For `f(x) = tan x = sin x/cos x`

`f'(x) = lim_(h->0) (tan(x+h) - tan x)/h`

= `lim_(h->0) ((sin(x+h))/(cos(x+h)) - sin x/cos x)/h`

= `lim_(h->0) (sin(x+h)*cos x - cos(x+h)*sin x)/(cos(x+h)*cos x*h)`

= `lim_(h->0) ((1/2)*sin(2x + h) + (1/2)*sin h - (1/2)*sin(2x + h) + (1/2)*sin h)/(cos(x+h)*cos x*h)`

= `lim_(h->0) ((sin h)/h)*(1/(cos(x+h)*cos x))`

When `h->0` , `sin h/h -> 1`

= `1/cos^2x`

= `sec^2x`

This proves from first principles that the derivative of tan x is sec^2x

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