How do you prove that the derivative of inverse tangent is 1x2+1?

Answer 1

Greetings! I present to you the proof to the derivative of inverse tangent. First, we let the inverse tangent be equal to y as a function of x. So, y = tan^-1x. Then, we take the tangent of both sides namely,

y = tan^-1x

tany = tan(tan^-1x)

tany = x. (Since the tangent of inverse tangent cancels out)

Now we compute implicit differentiation to both sides to obtain,

d/dx (tany) = d/dx (x)

sec²y (dy/dx) = 1.

Then divide sec²y to both sides to obtain,

dy/dx = 1 / sec²y.

Then use the trig identity to switch sec²y = 1 + (tany)²

dy/dx = 1 / 1+(tany)². Since x = tany,

dy / dx = 1 / 1+(x)².

-Completed

Answer 2

The best approach for solving this problem is to start with the tan(y) = x, and the use the Chain Rule together with Implicit differentiation to find the desired result. The procedure is detailed in the following image taken from a MathCad page which I created:

So the answer is gotten by substituting x for tan(y).

Answer 3

Let y = tan^-1(x)

→ tan(y) = x

By implicitly differentiating, we get

→ sec²(y) * y' = 1

→ y' = 1/[sec²(y)]

→ y' = cos²(y)

Now, going back to

tan(y) = x = x / 1, set up a right triangle with y as one of the acute angles,

and label the opposite side x, and the adjacent side as 1.

Then cos(y) = 1 / √(1 + x²)

→ cos²(y) = 1 / (1 + x²)

→ y' = 1 / (1 + x²)

Thank you for the question.

Michael Ehlers

Answer 4

Hello Kandace,

Let y = tan^-1(x)

→ tan(y) = x

By implicitly differentiating, we get

→ sec²(y) * y' = 1

→ y' = 1/[sec²(y)]

→ y' = cos²(y)

Now, going back to

tan(y) = x = x / 1, set up a right triangle with y as one of the acute angles,

and label the opposite side x, and the adjacent side as 1.

Then cos(y) = 1 / √(1 + x²)

→ cos²(y) = 1 / (1 + x²)

→ y' = 1 / (1 + x²)

Thank you for the question.

Michael Ehlers