Lagrange (Prime) Notation Examples

Find the derivative of e^4x.

Let u = 4x. Then we can rewrite:

e^4x = e^u

One of the patterns says

(e^u)' = e^u ⋅ u'.

Since u = 4x, we know u' = 4. We substitute these into the pattern and get

(e^4x)' = (e^u)' = e^u ⋅ u' = e^4x ⋅ 4.

We can also use these patterns to find antiderivatives. The equation

(e^u)' = e^u ⋅ u'

means the derivative of e^u is e^u ⋅ u'. Thinking backwards, this means

e^u is an antiderivative of e^u ⋅ u'.

We can write the family of all antiderivatives of e^u ⋅ u' as

e^u + C.

Using indefinite integral notation,

Find .

The coefficient 4 is the derivative of the exponent 4x:

4e^4x.

Let u = 4x. Then u' = 4. We can rewrite the integral using u:

Since multiplication is commutative,

Applying the pattern,

Finally, we put 4x back in for u to get

e^4x + C.

We conclude that

Rewriting the other patterns (assume a is a constant), we get

We can use any of these patterns to find indefinite integrals.

Find .

Let

u = 5x².

Then

u' = 10x.

We'll be using the pattern

We substitute u for 5x², apply the pattern, then put 5x² back in for u:

In the example  we didn't choose u = x². We chose u = 5x² because it's more complicated, but still "inside" the cosine function:

• u' (or some constant multiple of u') should already be a factor of the integrand.

In the example  we chose u = 5x². Its derivative, 10x, was already a factor of the integrand:

If the integral were

instead (without the factor x), we couldn't choose u = 5x² because its derivative u' = 10x isn't a factor of the integrand.

If the integral were

instead (without the factor 10), we could choose u = 5x² because the integrand includes the factor x, which is a constant multiple of u' = 10x.

For the integral, (a) identify u and u' and (b) integrate by substitution.

(a) We want u to be the most complicated thing that is still an "inside" function, so let

u = 5x
u' = 5

(b) Let's go through the three steps. Change variables:

Integrate using the appropriate pattern:

and finally put the original 5x back in for u:

-cos(u) + C = -cos(5x) + C.

We have

For the integral, (a) identify u and u' and (b) integrate by substitution.

(a) The most complicated "inside" function is 3x + 2, so let

u = 3x + 2.

Then

u' = 3.

(b) To change variables, we break up the coefficient 15 into 5 ⋅ 3 so we can see where u' is:

We undo the power rule to find the integral:

and put 3x + 2 back in for u:

(u)⁵ + C = (3x + 2)⁵ + C

Our final answer is

For the integral, (a) identify u and u' and (b) integrate by substitution.

(a) The most complicated "inside" function is the exponent of e:

We let

u = 4x²

u' = 8x

(b) We change variables, integrate using an appropriate pattern, and put the original variable back in:

Find .

We have

u = (x² + 4)

u' = 2x.

The integrand does have the factor 2x hiding in it, as we can see if we break up the 16 into 8 ⋅ 2:

From here we change variables, integrate, and put the original variable back in:

Sometimes, however, we really do need to introduce a new factor to the integrand. To do this we use the trick of multiplying by a clever form of 1.

Find .

We choose u = (x² + 4). Then u' = 2x. We have the factor x in the integrand, but we're missing the 2.

It's time for a trick. Multiply the integrand by

Since , we're not changing the value of the integrand.

Now we do have 2x as a factor of the integrand, so we can change variables:

Using one of the properties of integrals we can pull the constant  out of the integral. Then we can integrate and put the original variable back in:

As our final answer we write

You might reasonably wonder how the  turned into just + C.

The expression

accounts for the entire family of antiderivatives of (x² + 4)⁷x. So does the expression

Since both these expressions describe the same collection of functions, it doesn't matter which one we use. So we might as well use the simpler one and just write + C.

Find .

Let u = 6x + 3. Then u' = 6. We don't have a factor of 6 in the integrand, but we can get one. First we multiply the integrand by , which is a clever form of 1:

We pull the factor  out in front of the integrand and leave the factor 6 inside:

Now we can change variables and integrate:

Similarly to the previous example, we write + C instead of  because either way we're describing the same family of functions.

Here's a case where we can't use integration by substitution.

Find .

The most reasonable thing to try is to let u = x² + 4. Then u' = 2x. We have the factor 2 in the integrand. We might try to put a factor x in the integrand by multiplying by :

Now, however, we are stuck. We can't "pull out" the factor  to the outside of the integral. This doesn't work:

We can move constants back and forth across the integral sign, but not the variable of integration.

When multiplying by a clever form of 1, we can take a slight shortcut. What we've been doing is multiplying the integrand by a clever form of 1 and then moving one factor outside the integral:

Instead, we can immediately put the factor we want in the integrand inside the integral, and its reciprocal outside the integral: