Some of the derivatives of trigonometric function are shown below:

= d / dy sin (y) = cos (y);

= d / dy cos (y) = - sin (y);

= d / dy cot (y) = - csc

^{2}(y);

= d / dy ln (x) = 1 / x;

These all are different types of Derivatives of Trigonometric Functions.

Now we will see the derivative of lnx.

As we know, the derivative of lnx is 1 / x;

Let’s see prove of derivative of lnx.

First we write the ln x in the derivative form:

Proof = d / dx ln x = 1 /x;

We have to solve this derivative function by limit method.

In the limit form we can write it as:

= d / dx ln (x) = lim

_{(a > 0)}[ln (x + a) – ln (x)] / a = lim ln ((x + a) / x) / a;

In simple way we can write it as:

= lim (1 / a) ln (1 + a / x) = lim [ln (1 + a / x)

^{(1 / a)}];

Now let u = a / x and substitute the values then we get:

= lim

_{(a > 0)}[ln (1 + a)

^{(1 / ax)}] = 1 / x ln [lim

_{(a > 0)}(1 + a)

^{(1 / a)}];

If we further solve then we get:

= 1 / x ln (e); by the definition of e;

So we can write it as:

= 1 / x.

This is how to prove the derivative of lnx.