Chain Rule - Solved Examples

Example 1

Differentiate the following function by the help of the chain rule:

f(x) = sin(cos(x))

Solution:

Step 1: Identify the inner and outer functions

• Inner function is cos(x)
• Outer function is sin(cos(x))

Step 2: Differentiate the inner and outer functions separately
Inner function:

d/dx cos(x) = -sin(x)

Outer function:
Substitute a single variable in place of the inner function. Let u = cos(x), then

sin(cos(x)) = sin(u)

Differentiate the outer function with respect to the variable taken, u

d/dx sin(u) = cos(u)

Substitute back u = cos(x)

d/dx sin(u) = cos(cos(x))

Step 3: Multiply the derivatives of the inner and outer functions to get the derivative
f`(x) = -sin(x) * cos(cos(x)) = -sin(x)cos(cos(x))

The derivative of sin(cos(x)) is -sin(x)cos(cos(x))

Example 2

Differentiate the following function with the help of the chain rule:

f(x) = 2^sin(x)

Solution:

Identify the inner and outer functions. Here the inner function is sin(x) and outer function is an exponential function 2^sin(x). But we have a formula to differentiate the exponential functions, and it works only when there is a single variable in the exponent of a constant, so we will try to get a single variable instead of sin(x).

Substituting u = sin(x) we get 2^u. Now applying the chain rule formula:

d/dx f(g(x)) = d/du f(u) * d/dx u ... where u = g(x)

d/dx 2^sin(x) = d/du 2^u * d/dx u ... where u = sin(x)

Applying the derivative formula for exponential functions, derivative of 2^u =  2^u

d/dx 2^sin(x) = 2^u * d/dx u ... where u = sin(x)

Substituting u = sin(x),

d/dx 2^sin(x) = 2^sin(x) * d/dx sin(x)

The cosine function is the derivative of the sine function,

d/dx 2^sin(x) = 2^sin(x) * cos(x)

Therefore the derivative of  f(x) = 2^sin(x) is f `(x) = 2^sin(x) * cos(x)

An other, simpler way of doing the above steps without remembering the formula is to differentiate the inner and outer functions separately and then multiply their derivatives:

• Derivative of inner function, d/dx sin(x) = cos(x)
• Derivative of outer function, d/dx 2^u = 2^u
• Product of the two derivatives: 2^u * cos(x)
• Substitute back u = sin(x), 2^sin(x) * cos(x)

Example 3

Differentiate the following function with the help of the chain rule:

f(x) = (x² + 4x)⁴

Solution:

Method 1:

Identify the inner and outer functions.

• Inner function: x² + 4x
• Outer function: (x² + 4x)⁴

Substitute u = inner function so you get

Outer function: u⁴

Take the derivative of the inner and outer functions separately

• Inner function: d/dx (x² + 4x) = 2x + 4
• Outer function: d/du u⁴ = 4u³

Multiply the two derivatives and substitute back the inner function for u

f `(x) = (2x + 4) * 4u³

f `(x) = (2x + 4) * 4(x² + 4x)³

That is the derivative

Method 2:

After substituting u for the inner function, apply the chain rule formula:

d/dx f(g(x)) = d/du f(u) * d/dx g(x) where u = g(x)

d/dx (x² + 4x)⁴ = d/du u⁴ * d/dx (x² + 4x)

Apply the differentiation formulas and substitute the inner function back for u,

f `(x) = 4u³ * (2x + 4) 

f `(x) = 4(x² + 4x)³ * (2x + 4)

Example 4

Differentiate the following function by the help of the chain rule:

f(x) = ln(2x² + 3x + 1) 

Solution: 

Method 1:

Identify the inner and outer function:

• Inner function:  2x² + 3x + 1
• Outer function: ln(2x² + 3x + 1)

Substitute u for the inner function to get a single variable inside the outer function:

ln(2x² + 3x + 1) = ln(u), u = 2x² + 3x + 1

Differentiate the inner and outer functions separately:

• Inner function:  d/dx (2x² + 3x + 1) = 4x + 3
• Outer function: d/du ln(u) = 1/u

Substitute back the inner function for u

1/u = 1/(2x² + 3x + 1)

Multiply the two derivatives to get the derivative of f(x),

f `(x) = (4x + 3) * 1/(2x² + 3x + 1)

Simplify the expression as much as possible

f `(x) = (4x + 3)/(2x² + 3x + 1) 

Therefore the derivative of ln(2x² + 3x + 1) is (4x + 3)/(2x² + 3x + 1).

Method 2:

Identify the inner and outer functions. The inner function is (2x² + 3x + 1) and the outer one is the logarithmic function. Substitute u = inner function so you get the following. Remember that 'u' is not variable but a function, that is the inner function.

log(2x² + 3x + 1) = log(u)

Applying the chain rule formula,

d/dx f(g(x)) = d/du f(u) * d/dx g(x) where u = g(x)

d/dx log(2x² + 3x + 1) = d/du log(u) * d/dx (2x² + 3x + 1)

Compute the derivatives,

1/u * (4x + 3)

Substitute back the inner function for u,

1/(2x² + 3x + 1)  * (4x + 3)

Simplify,

(4x + 3)/(2x² + 3x + 1).

Example 5

Differentiate the following function with the help of the chain rule:

f(x) = √x² + 4x - 3

Solution:

Identify the inner and outer functions:

• Inner function: x² + 4x - 3
• Outer function:  √x² + 4x - 3

Substitute u = inner function in the outer function,

√x² + 4x - 3 = √u

Differentiate the inner and outer functions separately,

• Inner function: d/dx x² + 4x - 3 = 2x + 4
• Outer function: d/du √u = d/du u^1/2 = 1/2 u^-1/2

Multiply the two derivatives and substitute the inner function back for u,

(2x + 4) * (1/2 u^-1/2) = (2x + 4) * (1/2(x² + 4x - 3)^-1/2)

Simplify the expression,

(2x + 4)/(2√x² + 4x - 3 ) 

Therefore the derivative of √x² + 4x - 3 is (2x + 4)/(2√x² + 4x - 3 )

Example 6

Differentiate the following function with the help of the chain rule.

f(x) = sin(cos(tan(x)))

Solution:

Identify the inner and outer functions.

• Inner function: cos(tan(x)) 
• Outer function: sin(cos(tan(x))) 

Substitute u for the inner function

• Inner function: u = cos(tan(x)) 
• Outer function: sin(u) 

Apply the chain rule formula,

d/dx f(g(x)) = d/du f(u) * d/dx g(x) ... where u = g(x),

d/dx sin(cos(tan(x))) = d/du sin(u) * d/dx cos(tan(x))

Derivative of sin(u) is cos(u),

 = cos(u) * d/dx cos(tan(x))

Substitute back u = cos(tan(x)),

 = cos(cos(tan(x))) * d/dx cos(tan(x))

Use chain rule again on the second derivative because it is a composition of two functions. Identify the inner and outer functions,

• Inner function: tan(x) 
• Outer function: cos(tan(x)) 

Substitute u for the inner function

• Inner function: u = tan(x) 
• Outer function: cos(u) 

Apply the chain rule formula,

d/dx f(g(x)) = d/du f(u) * d/dx g(x) ... where u = g(x), 

d/dx cos(tan(x)) = d/du cos(u) * d/dx tan(x)

Derivative of cos(u) is -sin(u) and that of tan(x) is sec2(x)

 = -sin(u) * sec²(x)

Substitute back u = tan(x),

 = -sin(tan(x)) * sec²(x)

Substitute this derivative of cos(tan(x)) in the derivative obtained after the first application of the chain rule,

 = cos(cos(tan(x))) * d/dx cos(tan(x)) 

 = cos(cos(tan(x))) * -sin(tan(x)) * sec²(x)

Simplify the expression,

 = -sin(tan(x)) cos(cos(tan(x))) sec²(x)