If f(x) = ln(x^2 + 1), what is f′(x)? Step by Step

Step by step

The question

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

y = ln(x² + 1)

There are four choices.

First we solve it, the normal way.

Then we pick.

1 The inside

There is something inside: x² + 1.

Call it u.

u = x² + 1

Then y is just ln u.

y = ln u

2 The outside

The derivative of ln x is 1 over x.

ln x → 1x

First the outside.

The rule on ln u gives 1 over u.

dydu = 1u

3 The inside's slope

Now the inside.

The power rule on x² + 1 gives 2x.

dudx = 2x

4 Multiply

The chain rule: the outside's derivative times the inside's derivative, dy/du times du/dx.

dydx = dydu × dudx

Put the inside back in place of u.

= 1x² + 1 × 2x

Careful!

Do not stop here!

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

Careful: not 1x² + 1 ✗

So dy/dx = 2x/(x² + 1).

= 2xx² + 1

5 Check

Let's check: the derivative is the slope.

At x = 1, the rule gives 1, with a calculator.

dy/dx at x = 1: 1

Now the slope between two very close points: x = 1 and x = 1.001.

A calculator gives about 1.

Very close to 1.

Right!

y(1.001) − y(1)0.001 ≈ 1

6 Pick the choice

Our answer is dy over dx = 2x/(x² + 1).

That is choice B.

B) 2xx² + 1

7 Why not the others

A is 1/(x² + 1).

Wrong: lost × 2x, the derivative of the inside.

That is the Careful mistake from before.

A) 1x² + 1 ✗ lost × 2x, the derivative of the inside

C is 2x.

Wrong: ln u gives u′ over u: the bottom stays.

C) 2x ✗ ln u gives u′ over u: the bottom stays

D is 1/(2x).

Wrong: took 1 over the inside's derivative.

D) 12x ✗ took 1 over the inside's derivative

Let's look back.

First: the inside.

Then: the outside.

Then: the inside's slope.

Then: multiply.

Then: check.

Then: pick the choice.

Last: why not the others.

Answer