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.