# Calculus Chain Rule

Chain rule is a formula for solving the derivative of a composite of two functions.

The Composite function u o v of functions u and v is the function whose values  u[v(x)] are found for each x in the domain of v for which v(x) is in the domain of u.

## Chain Rule

If the function v can be differentiated at x, and the function u can be differentiated at  v(x) , then the composite function uov can be differentiated at x.

The value of the derivative of uov at x is given as follows:

d/dx(uov) (x) =u’[v(x)] . (dv)/dx

d/dx[(uov)(x)] =(du/dv) . (dv/dx)

We now present a few examples that will help you understand how to find the derivative of a composite function using the Chain Rule.

## Find the derivative of the function using Chain Rule: y = (x^2 + 2)^5

#### Explanation:

Let u =  x^2 + 2  => y = u^5

Using the Chain Rule stated above:

dy/dx = dy/du . du/dx

We have in this example:

dy/du= d/du(u^5) = 5u^4

And

 du/dx = d/dx( x^2+ 2 ) = 2x

=> dy/dx = (5u^4). (2x) = (5(x^2 + 2)^4) . (2x)

=> dy/dx = 10x(x^2 + 2)^4

that is the required derivative to the given function.

## Find the derivative dy/dx of the function using Chain Rule given in Parametric Equation form: x = ct^3 y = 4ct

 ( 64c^2)/( 3y^2 )

#### Explanation:

Since the two variables appear in two different equations in the variable t, we can make use of the Chain Rule in order to find the derivative of y w.r.t. x,

i.e. dy/dx

According to the Chain Rule Formula:

dy/dx = dy/dt . dt/dx  ( Heredt/dx= 1/(dx/dt) )

We have

dy/dt= d/dt(4ct ) = 4c  and dx/dt = d/dt (ct^3) = 3ct^2

Now

dy/dx  =  dy/dt . dt/dx = (dy/dt)/(dx/dt) =( 4c )/( 3ct^2 ) = ( 4 )/( 3t^2 )

We have

t =y/4c => t2 = y^2/(16c^2 )

Hence

=> dy/dx =( 4 )/( 3 y^2/(16c^2 )) = ( 64c^2)/( 3y^2 )

that is the desired derivative dy/dx of the given function.

## Derivatives of Polynomial Functions

A polynomial is nothing but a combination of expressions of the form  1, x, x^2, x^3…. x^n  .By making use of the power Formula:  (d [x^n])/dx= n. xn – 1 ,  and the theorem of summation, we can find the derivative of any polynomial function.

Presented below is an example of finding the derivative of a polynomial function.

## Find the derivative  dy/dx  of the polynomial function :  y = 4x^4 + 5x^3- 6x^2 + 2x + 2

#### Explanation:

We use the power formula to find the derivative of the individual terms that make up the polynomial expression as shown below.

 dy/dx = ( d )/(dx )(4x^4 + 5x^3- 6x2 + 2x + 2)

 dy/dx = ( d )/(dx )(4x^4) + ( d )/(dx )(5x^3)-( d )/(dx )(6x^2) + ( d )/(dx )(2x) + ( d )/(dx )(2)

 dy/dx = 4. (4x^3) + 5 (3x^2) - 6(2x) + 2 + 0

 dy/dx = 16x^3 + 15x^2- 12x + 2

that is the required derivative of the given polynomial function in x.

## Derivative of Trigonometric Functions

Note: While finding derivatives of trigonometric functions, we assume that x, the variable of differentiation, is measured in radians.
For brevity, we enlist the derivatives of the six fundamental trigonometric functions below:

•   d/dx(sin x) = cos x
•   d/dx(cos x) = -sin x
•   d/dx(tan x) = sec2 x
•   d/dx(cot x) = -cosec2 x
•   d/dx(cosec x) = - cosec x cot x
•   d/dx(sec x) = sec x tan x

All these identities are very simple and straight forward. And you can employ the very basic first principle / ab-initio / by definition method to derive these results.

Two basic formulas that have been used in the derivation of these results have been given below:

Lim_(x->0) ( sin x )/x = 1

 Lim_(x->0) ( 1 - cos x)/x = 0

We now present some examples that make use of these identities and also include some interesting concepts regarding the solving techniques.

## Find the derivative of the function w.r.t. x.:  cos 3x

#### Explanation:

Let

 y = cos 3x  and  u = 3x => y = cos u

Using the chain Rule Formula:

 dy/dx=dy/du . du/dx

We have

 dy/du=d/du( cos u ) = - sin u

And

 du/dx=d/dx(3x) = 3

Hence

 dy/dx=dy/du . du/dx= (- sin u ) . 3 = -3sin u

Putting the value of u into the above expression gives:

 dy/dx= -3sin 3x

Which is the desired derivative of the given trigonometric function cos 3x

## Find the derivative of the function w.r.t. x. :  cot2 x

#### Explanation:

Let  y = cot2x  and  u = cot x => y = u2

Using the chain Rule Formula:

 dy/dx=dy/du . du/dx

We have

 dy/du=d/du(u2) = 2u

And

 du/dx=d/dx(cot x) = -cosec2 x

Hence

 dy/dx=dy/du . du/dx= (2u ) .( -cosec2 x) = -2ucosec2 x

Putting the value of u into the above expression gives us the required derivative.

 dy/dx= -2cot xcosec2 x

## Differentiate  sin3x  w.r.t.  cos2x

 ( -3 )/2sinx

#### Explanation:

Let  sin3x = u  and  cos2x = v

Then we are supposed to find du/dv . We shall use Chain Rule along with the identities of trigonometric functions stated above.

 du/dx= d/(dx )(sin3x) = 3sin2x.cos x and dv/dx = d/(dx )(cos2x) = 2cos x . (-sin x)

 => du/dv= ( du)/( dx ) .( dx )/( dv )= ( 3sin^2 x . cos x)/(-2 cos?x .sin?x )= ( -3 )/2sinx

## Derivatives of Inverse Trigonometric Functions

Presented below is the list of the derivatives of Inverse Trigonometric Functions. Simple submissions can be used to prove these identities (For brevity, their proof is not being included in this text).

•   d/dx(Sin-1 x) = ( 1 )/v(1- x^2 )
•   d/dx(Cos-1 x) = -( 1 )/v(1- x^2 )
•   d/dx(Tan-1 x) = ( 1 )/( 1 + x^(2 ) )
•    d/dx(Cot-1 x) = -( 1 )/( 1 + x^(2 ) )
•   d/dx(Cosec-1 x) =-( 1 )/(|x| v(x^2- 1))
•   d/dx(Sec-1 x) = ( 1 )/(|x| v(x^2- 1))

## Find the derivative  dy/dx of the following function.  y = tan( 2 Tan-1 x/2)

( 4 ( 1 + y^(2 )) )/( 4 + x^(2 ) )

#### Explanation:

Let  u = 2 Tan-1 x/2 => y= tan u

 =>dy/du= sec2 u = 1 + tan2u = 1 + y2

And  du/dx=d/dx(2 Tan-1 x/2)=2 . ( 1 )/( 1 + ( x/( 2 ) )^(2 ) ). d/dx (x/( 2 ))

 =2 .( 1 )/( 1 + x^(2 )/4) .1/( 2 ) = ( 4 )/( 4 + x^(2 ) )

Using the Chain Rule,

 dy/dx = dy/du . du/dx = ( 1 + y2 ) . ( 4 )/( 4 + x^(2 ) )= ( 4 ( 1 + y^(2 )) )/( 4 + x^(2 ) )

that is the desired derivative  dy/dx

## Derivatives of Logarithmic and Exponential Functions

An exponential function is of the form

 f(x) = ax where  a > 0 , a ? 1 and x is any real number.

A logarithmic function is of the form

y =  loga x (x>0)  where  x = a y , a > 0  and  a ? 1

is called the logarithm of x to the base a.

Some basic formulas associated with the derivatives of logarithmic and exponential functions are as follows:

 d/dx( e^x) = e^x

 d/dx( a^x) = a^x.  ( lna ) (where ln is the natural logarithm: log with base e)

 d/dx(lnx) = 1/x

 d/dx(loga x) = 1/x . 1/(ln a)

We next present a few examples of finding derivatives of logarithmic and exponential functions.

## Find the derivative dy/dx of the function:  y = e^(x^2+ 1)

 2x . e^(x^2+ 1)

#### Explanation:

Given that  y = e^(x^2+ 1)

Putting  u = x^2+1 => y = eu

We have  dy/du= eu => dy/du =e^(x^2+ 1)

And  du/dx=d/dx( x^2+1 )= 2x

Using the Chain Rule Formula  dy/dx = dy/du . du/dx= ( e^(x^2+ 1)). ( 2x ) = 2x . e^(x^2+ 1)

That is the desired derivative of the given function.

## Find the derivative dy/dx of the following function.  y = ln (x^2 + 2x)

 (2 ( x + 1 ))/( x^2+ 2x )

#### Explanation:

Let  u = x2 + 2x => du/dx=d/dx( x^2+2x )= 2x + 2

=> y = ln u => dy/du=d/du( ln u ) = 1/u= 1/( x^2+ 2x )

Using the Chain Rule Formula

 dy/dx = dy/du . du/dx =1/( x^2+ 2x ) . 2x + 2 = (2 ( x + 1 ))/( x^2+ 2x )

That is the desired derivative of the given function.

## Higher Order Derivatives

If we have f’  as the first derivative of a function  f, then  (f’)’  is the derivative of  f’ , and is called the second derivative of  f.  Similarly we have higher order derivatives such as 3rd, 4th, 5th derivatives and so on.

There is nothing new about the concept of higher derivative. A second derivative is the for  f’ as  f’ is to  f. And also the rules for computing higher derivatives are same.

We present some examples of higher derivatives next.

## Find higher derivatives of the polynomial.  f(x) = 1/( 12)x^4–( 1 )/6x^3 +( 1 )/( 4 )x^2 + 2x + 7

#### Explanation:

 f ’(x) = 1/( 12)( 4x^3 ) –( 1 )/6( 3x^2 ) + ( 1 )/( 4 )( 2x) + 2 + 0 = 1/( 3 )x^3 –( 1 )/2 x^2 + ( 1 )/2 x + 2

 f ’’(x) = 1/( 3 )(3x^2)–( 1 )/2 2x + ( 1 )/2 + 0 = x^2– x + ( 1 )/2

 f ”’(x) = 2x –1

 f^iv (x) = 2

And all other higher derivatives are zero.
Conclusion: The number of non-zero derivatives of a polynomial of degree n is n.

#### Become a member today (it’s Free)!

Register with us