^{2}(x). Now we will see the prove of the derivative of Tan Squared X.

To find the derivative of Tan Squared X first we have to write the function tan

^{2}x in the derivative form:

So we can write it as:

= $\frac{d}{dx}$ tan

^{2}x = 2 tan x sec

^{2}x;

We known that the derivative of tan x is sec

^{2}x but we solve it using the Chain Rule.

Now use the chain rule, $\frac{d}{dx}$ (tan

^{2}(x)) =

$\frac{d}{dx}$ u

, where the value of u = tan (x) and

$\frac{du}{dx}$ = sec

^{2}x

2u;

So put these values in the trigonometric function:

On putting these values we get:

= X (2 tan (x) ($\frac{du}{dx}$ (tan (x)));

We already know that if we differentiate tan x then we get sec

^{2}x, so put sec

^{2}x in the above function.

= X (2 tan (x) (sec

^{2}(x)));

After differentiating the trigonometric function we get:

= 2 X tan x sec

^{2}(x);

This is how one can solve the derivative of derivative Of Tan Squared X. Let’s see an example to solve the derivative of other trigonometric function. Suppose a function is given f (s) = s

^{2}sin (6x); then we have to find the inverse derivative of a given function.

By using Product rule we get.

u.v = u $\frac{du}{dx}$ v + v. $\frac{du}{dx}$ u;

So we can write as:

$\frac{d}{ds}$ f (s) = $\frac{d}{ds}$ (s

^{2}sin 6x)

= sin 6x $\frac{d}{ds}$ (s

^{2})

(sin (6s));

If we differentiate the value s

^{2}then we get 2s and we differentiate sin 6s then we get 6 cos 6s;

Now putting these values in the given function we get:

f (s) = 2s.sin (6s) +s

^{2}. 6(cos (6s));

We can also write it as:

f (s) = 2s.sin (6s) + 6s

^{2}6(cos (6s));

This is how to calculate the inverse Derivatives of trigonometric function.