Derivative of 1/(x^2 + 1)^2 Step by Step

Step by step

The question

Question: find dy/dx for y = 1/(x² + 1)².

y = 1(x² + 1)²

1 The inside

There is something inside: x² + 1.

Call it u.

u = x² + 1

Then y is just u⁻².

Written as a power of u.

y = u⁻²

u to a power at the bottom is a minus power on top.

1u² = u⁻²

2 The outside

First the outside.

Use the power rule on u⁻².

dydu = (−2)u^(−2 − 1)

We get −2 u⁻³.

= −2u⁻³

Take 1 away from the power: −2 − 1 = −3.

−2 − 1 = −3

3 The inside's slope

Now the inside.

The power rule on x² + 1 gives 2x.

dudx = 2x

4 Multiply

The chain rule: multiply the two.

dy/du times du/dx.

dydx = dydu × dudx

Put the inside back in place of u.

= −2(x² + 1)⁻³ × 2x

Careful!

Do not stop here!

We must multiply by the inside's slope, 2x.

Careful: not −2(x² + 1)⁻³ ✗

Multiply the numbers in front.

Then write the minus power at the bottom.

So dy/dx = −4x/(x² + 1)³.

= −4x(x² + 1)³

−2 × 2x = −4x

5 Check

Let's check: the derivative is the slope.

At x = 0, the rule gives 0.

dy/dx at x = 0: 0

Now the slope between two very close points: x = 0 and x = 0.001.

A calculator gives about 0.

Very close to 0.

Right!

y(0.001) − y(0)0.001 ≈ 0

Let's look back.

First: the inside.

Then: the outside.

Then: the inside's slope.

Then: multiply.

Last: check.

Answer