Demostrate that:

d/dx [tan^-1(x)]= 1/1+ x^2

-> using implicit differentiation.

To demonstrate this using implicit differentiation, we will consider the equation:

y = tan^(-1)(x)

Differentiating both sides with respect to x will give us:

dy/dx = d/dx[tan^(-1)(x)]

Now, let's focus on the right side of the equation and differentiate it using implicit differentiation.

Using the chain rule, we have:

d/dx[tan^(-1)(x)] = d/dx[tan^(-1)(y)] * dy/dx

First, let's find d/dx[tan^(-1)(y)]:

We know that tan^(-1)(tan(x)) = x, for any value of x.

Applying the inverse trigonometric identity, we rewrite this as:

tan(tan^(-1)(y)) = y

Differentiating both sides with respect to x, we get:

sec^2(tan^(-1)(y)) * d/dx[tan^(-1)(y)] = dy/dx

Now, note that sec^2(tan^(-1)(y)) can be rewritten as 1 + tan^2(tan^(-1)(y)).

Using the trigonometric identity, we have:

1 + tan^2(tan^(-1)(y)) = 1 + y^2

Substituting this into the previous equation, we get:

(1 + y^2) * d/dx[tan^(-1)(y)] = dy/dx

Now, let's substitute y = tan^(-1)(x) back into the equation:

(1 + (tan^(-1)(x))^2) * d/dx[tan^(-1)(x)] = dy/dx

Since dy/dx = d/dx[tan^(-1)(x)], we can simplify further:

(1 + (tan^(-1)(x))^2) * d/dx[tan^(-1)(x)] = d/dx[tan^(-1)(x)]

Dividing both sides of the equation by (1 + (tan^(-1)(x))^2), we have:

d/dx[tan^(-1)(x)] = 1 / (1 + (tan^(-1)(x))^2)

Thus, we have proven that d/dx[tan^(-1)(x)] = 1 / (1 + x^2) using implicit differentiation.