3.6 The Chain Rule - Dr. Travers Page of...

§ 3.6 The Chain Rule

The Rule

Ruledydx

=dydu· du

dx

or

Ruleddx

(f ◦ g)(x) =ddx

f (g(x)) = f ′(g(x)) · g′(x)

or

Ruleddx

f (u) = f ′(u)dudx

The Rule

Ruledydx

=dydu· du

dx

or

Ruleddx

(f ◦ g)(x) =ddx

f (g(x)) = f ′(g(x)) · g′(x)

or

Ruleddx


The Rule

Ruledydx

=dydu· du

dx

or

Ruleddx

(f ◦ g)(x) =ddx

f (g(x)) = f ′(g(x)) · g′(x)

or

Ruleddx


What This Means

When we have a composite function, we first identify the inside andoutside functions.

Then we find the derivative of the outside function but leave theinside function as it’s variable.

Then we multiply this result by the derivative of the inside function.

What This Means




What This Means




Example 1

Example

Find y′ if y = (4x2 + 1)7.

What is the inside function?u = 4x2 + 1What is the derivative?u′ = 8x

What is the outside function?y = u7

What is the derivative?y′ = 7u6

Putting this together, we have

dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.

What is the inside function?

u = 4x2 + 1What is the derivative?u′ = 8x




dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.

What is the inside function?u = 4x2 + 1

What is the derivative?u′ = 8x




dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.

What is the inside function?u = 4x2 + 1What is the derivative?

u′ = 8x




dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.





dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.


What is the outside function?

y = u7



dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.





dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.



What is the derivative?

y′ = 7u6


dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.





dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.





dydx

=dydu· du

dx

= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.





dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.





dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 1

Example

Find y′ if y = (4x2 + 1)7.





dydx

=dydu· du

dx= 7u6 · 8x

= 7(4x2 + 1)6 · 8x

= 56x(4x2 + 1)6

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.

What is the inside function?u = 3x2 + 5x− 2What is the derivative?u′ = 6x + 5

What is the outside function?y =√

uWhat is the derivative?y′ = 1

2√

uPutting this together, we have

dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.


u = 3x2 + 5x− 2What is the derivative?u′ = 6x + 5



2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.

What is the inside function?u = 3x2 + 5x− 2

What is the derivative?u′ = 6x + 5



2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.

What is the inside function?u = 3x2 + 5x− 2What is the derivative?

u′ = 6x + 5



2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.




2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.



y =√


2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.



u

What is the derivative?y′ = 1

2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.



uWhat is the derivative?

y′ = 12√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.




2√

u


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.




2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.




2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 2

Example

Find y′ if y =√

3x2 + 5x− 2.




2√


dydx

=dydu· du

dx

=1

2√

u· (6x + 5)

=6x + 5

2√

3x2 + 5x− 2

Example 3

Example

Find y′ if y = 1x2+x4 .

What is the inside function?u = x2 + x4

What is the derivative?u′ = 2x + 4x3

What is the outside function?y = 1

uWhat is the derivative?y′ = − 1

u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example



u = x2 + x4




u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example






u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example




u′ = 2x + 4x3



u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example






u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example





y = 1u

What is the derivative?y′ = − 1

u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example





u

What is the derivative?y′ = − 1

u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example






y′ = − 1u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example






u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example






u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example






u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 3

Example






u2


dydx

=dydu· du

dx

= − 1u2 · (2x + 4x3)

= − 2x + 4x3

(x2 + x4)2

Example 4

Example

Find y′ if y = sin4 2x

What is the inside function?u = sin 2xWhat is the derivative?u′ = cos 2x · d

dx 2xu′ = 2 cos 2x




dydx

=dydu· du

dx= 4u3 · (2 cos 2x)

= 8 sin3 2x · cos 2x

Example 4

Example

Find y′ if y = sin4 2xWhat is the inside function?

u = sin 2xWhat is the derivative?u′ = cos 2x · d





dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example

Find y′ if y = sin4 2xWhat is the inside function?u = sin 2x

What is the derivative?u′ = cos 2x · d





dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example

Find y′ if y = sin4 2xWhat is the inside function?u = sin 2xWhat is the derivative?

u′ = cos 2x · ddx 2x

u′ = 2 cos 2x




dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example

Find y′ if y = sin4 2xWhat is the inside function?u = sin 2xWhat is the derivative?u′ = cos 2x

· ddx 2x

u′ = 2 cos 2x




dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example

Find y′ if y = sin4 2xWhat is the inside function?u = sin 2xWhat is the derivative?u′ = cos 2x · d

dx 2x

u′ = 2 cos 2x




dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example






dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example




y = u4



dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example






dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example





y′ = 4u3


dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example






dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example






dydx

=dydu· du

dx

= 4u3 · (2 cos 2x)


Example 4

Example






dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Example 4

Example






dydx

=dydu· du

dx= 4u3 · (2 cos 2x)


Exponential Functions

What do exponential functions look like when we graph them?

What do the equations of exponential functions look like?

f (x) = cax

We will focus here on functions of the form f (x) = ax for a ∈ R,a > 0.




f (x) = cax





f (x) = cax





f (x) = cax



We want to find the derivative of f (x) = ax. What is the onlytechnique we know at this point?

ddx

ax = limh→0

ax+h − ax

h

= limh→0

ax(ah − 1)h

= ax limh→0

ah − 1h

Now what?

Let’s look at this numerically.



ddx

ax = limh→0

ax+h − ax

h

= limh→0

ax(ah − 1)h

= ax limh→0

ah − 1h

Now what?




ddx

ax = limh→0

ax+h − ax

h

= limh→0

ax(ah − 1)h

= ax limh→0

ah − 1h

Now what?




ddx

ax = limh→0

ax+h − ax

h

= limh→0

ax(ah − 1)h

= ax limh→0

ah − 1h

Now what?




ddx

ax = limh→0

ax+h − ax

h

= limh→0

ax(ah − 1)h

= ax limh→0

ah − 1h

Now what?




ddx

ax = limh→0

ax+h − ax

h

= limh→0

ax(ah − 1)h

= ax limh→0

ah − 1h

Now what?


Derivative of f (x) = 2x

Let’s look at the limit part, since we know that the ax part is notdependent on the limit.

h 2h−1h

.1 .7177.01 .695.001 .6934

Does this number look familiar to anyone?



h 2h−1h

.1

.7177.01 .695.001 .6934




h 2h−1h

.1 .7177

.01 .695.001 .6934




h 2h−1h

.1 .7177.01

.695.001 .6934




h 2h−1h

.1 .7177.01 .695

.001 .6934




h 2h−1h

.1 .7177.01 .695.001

.6934




h 2h−1h

.1 .7177.01 .695.001 .6934




h 2h−1h

.1 .7177.01 .695.001 .6934




h 2h−1h

.1 .7177.01 .695.001 .6934



Let’s try something similar for f (x) = 5x.

h 5h−1h

.1 1.746.01 1.622.001 1.611

What number are we approaching here? Is there a pattern?



h 5h−1h

.1

1.746.01 1.622.001 1.611




h 5h−1h

.1 1.746

.01 1.622.001 1.611




h 5h−1h

.1 1.746.01

1.622.001 1.611




h 5h−1h

.1 1.746.01 1.622

.001 1.611




h 5h−1h

.1 1.746.01 1.622.001

1.611




h 5h−1h

.1 1.746.01 1.622.001 1.611




h 5h−1h

.1 1.746.01 1.622.001 1.611




h 5h−1h

.1 1.746.01 1.622.001 1.611


Derivative of f (x) = ax

.7177, .6955, .6934⇒

ln 2

1.746, 1.622, 1.611⇒ ln 5

and this makes our derivative ...


ddx

ax = axln a


.7177, .6955, .6934⇒ ln 2

1.746, 1.622, 1.611⇒ ln 5



ddx

ax = axln a


.7177, .6955, .6934⇒ ln 2

1.746, 1.622, 1.611⇒

ln 5



ddx

ax = axln a


.7177, .6955, .6934⇒ ln 2

1.746, 1.622, 1.611⇒ ln 5



ddx

ax = axln a


.7177, .6955, .6934⇒ ln 2

1.746, 1.622, 1.611⇒ ln 5



ddx

ax = axln a


.7177, .6955, .6934⇒ ln 2

1.746, 1.622, 1.611⇒ ln 5



ddx

ax = axln a

Hmmm ...

Is there an a such that ddx ax = ax?

Using what we just did, we want limh→0

ah−1h = 1.

Consider the following:

ah − 1h≈ 1

ah − 1 ≈ h

ah ≈ h + 1

a ≈ (1 + h)1h

Hmmm ...



ah−1h = 1.


ah − 1h≈ 1

ah − 1 ≈ h

ah ≈ h + 1

a ≈ (1 + h)1h

Hmmm ...



ah−1h = 1.


ah − 1h≈ 1

ah − 1 ≈ h

ah ≈ h + 1

a ≈ (1 + h)1h

Hmmm ...



ah−1h = 1.


ah − 1h≈ 1

ah − 1 ≈ h

ah ≈ h + 1

a ≈ (1 + h)1h

Hmmm ...



ah−1h = 1.


ah − 1h≈ 1

ah − 1 ≈ h

ah ≈ h + 1

a ≈ (1 + h)1h

Hmmm ...



ah−1h = 1.


ah − 1h≈ 1

ah − 1 ≈ h

ah ≈ h + 1

a ≈ (1 + h)1h

Hmmm ... (cont.)

Now let’s bring back the limit. Does this look familiar?

limh→0

(1 + h)1h

This may look more familiar in another form ...

limh→∞

(1 +

1h

)h

= e

Derivative of f (x) = ex

ddx

ex = ex

Hmmm ... (cont.)


limh→0

(1 + h)1h


limh→∞

(1 +

1h

)h

=

e


ddx

ex = ex

Hmmm ... (cont.)


limh→0

(1 + h)1h


limh→∞

(1 +

1h

)h

= e


ddx

ex = ex

Hmmm ... (cont.)


limh→0

(1 + h)1h


limh→∞

(1 +

1h

)h

= e


ddx

ex = ex

Example 5

Example

Find f ′(x) if f (x) = e3x

What is the inside function?u = 3xWhat is the derivative?u′ = 3

What is the outside function?y = eu

What is the derivative?y′ = eu


f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example



u = 3xWhat is the derivative?u′ = 3




f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example


What is the inside function?u = 3x

What is the derivative?u′ = 3




f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example


What is the inside function?u = 3xWhat is the derivative?

u′ = 3




f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example






f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example




y = eu



f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example






f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example





y′ = eu


f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example






f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 5

Example






f ′(x) =dydu· du

dx

= eu · 3= 3 e3x

Example 5

Example






f ′(x) =dydu· du

dx= eu · 3

= 3 e3x

Example 5

Example






f ′(x) =dydu· du

dx= eu · 3= 3 e3x

Example 6

Example

Find f ′(x) for f (x) =√

ex + 1.

What is the inside function?u = ex + 1What is the derivative?u′ = ex



2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.


u = ex + 1What is the derivative?u′ = ex



2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.

What is the inside function?u = ex + 1

What is the derivative?u′ = ex



2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.

What is the inside function?u = ex + 1What is the derivative?

u′ = ex



2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.




2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.



y =√


2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.



u

What is the derivative?y′ = 1

2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.




y′ = 12√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.




2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.




2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.




2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 6

Example


ex + 1.




2√

u


f ′(x) =dydu· du

dx

=1

2√

u· ex

=ex

2√

ex + 1

Example 7

Example

Find f ′(x) for f (x) = ex2.

What is the inside function?u = x2





f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example



u = x2





f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example







f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example




u′ = 2x




f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example







f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example





y = eu



f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example







f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example






y′ = eu


f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example







f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example







f ′(x) =dydu· du

dx

= eu · 2x

= 2xex2

Example 7

Example







f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 7

Example







f ′(x) =dydu· du

dx= eu · 2x

= 2xex2

Example 8

Example

Find where f (x) = 2x · e3x+1 has a horizontal tangent line.

What do we need to do here?

We need to use the product rule and the chain rule and set thederivative equal to 0.

Example 8

Example




Example 8

Example




Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)

Can e3x+1 ever equal 0?

e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2

= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)


e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)


e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)


e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)


e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)


e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)


e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)


e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 8

f ′(x) = 2xddx

e3x+1 + e3x+1 ddx

2x

= 2x(3e3x+1)+ e3x+1 · 2= 6xe3x+1 + 2e3x+1

= e3x+1(6x + 2)


e3x+1(6x + 2) = 0

6x + 2 = 0

6x = −2

x = −13

Example 9

Example


(x2 · 5x)3.

ddx

√(x2 · 5x)3 =

1

2√(x2 · 5x)3

· ddx

(x2 · 5x)3

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · ddx

(x2 · 5x)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 ·(

x2 ddx

5x + 5x ddx

x2)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · (x2 · 5x ln 5 + 5x · 2x)

Example 9

Example


(x2 · 5x)3.

ddx

√(x2 · 5x)3

=1

2√(x2 · 5x)3

· ddx

(x2 · 5x)3

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · ddx

(x2 · 5x)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 ·(

x2 ddx

5x + 5x ddx

x2)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · (x2 · 5x ln 5 + 5x · 2x)

Example 9

Example


(x2 · 5x)3.

ddx

√(x2 · 5x)3 =

1

2√

(x2 · 5x)3· d

dx(x2 · 5x)3

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · ddx

(x2 · 5x)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 ·(

x2 ddx

5x + 5x ddx

x2)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · (x2 · 5x ln 5 + 5x · 2x)

Example 9

Example


(x2 · 5x)3.

ddx

√(x2 · 5x)3 =

1

2√

(x2 · 5x)3· d

dx(x2 · 5x)3

=1

2√

(x2 · 5x)3· 3(x2 · 5x)2 · d

dx(x2 · 5x)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 ·(

x2 ddx

5x + 5x ddx

x2)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · (x2 · 5x ln 5 + 5x · 2x)

Example 9

Example


(x2 · 5x)3.

ddx

√(x2 · 5x)3 =

1

2√

(x2 · 5x)3· d

dx(x2 · 5x)3

=1

2√

(x2 · 5x)3· 3(x2 · 5x)2 · d

dx(x2 · 5x)

=1

2√

(x2 · 5x)3· 3(x2 · 5x)2 ·

(x2 d

dx5x + 5x d

dxx2)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · (x2 · 5x ln 5 + 5x · 2x)

Example 9

Example


(x2 · 5x)3.

ddx

√(x2 · 5x)3 =

1

2√

(x2 · 5x)3· d

dx(x2 · 5x)3

=1

2√

(x2 · 5x)3· 3(x2 · 5x)2 · d

dx(x2 · 5x)

=1

2√

(x2 · 5x)3· 3(x2 · 5x)2 ·

(x2 d

dx5x + 5x d

dxx2)

=1

2√(x2 · 5x)3

· 3(x2 · 5x)2 · (x2 · 5x ln 5 + 5x · 2x)

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