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 eachxin the domain ofvfor which`v(x)`is in the domain ofu.

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**.

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`

=> `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.

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

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.

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.

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 `

` 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.

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 `

` 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.

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

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 `

` ( -3 )/2sinx `

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 `

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

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)) `

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

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 ) ) `

` =>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 `

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) `

` 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.

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

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)`

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.

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

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 ) `

`=> 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.

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.

` f(x) = 1/( 12)x^4–( 1 )/6x^3 +( 1 )/( 4 )x^2 + 2x + 7 `

` 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`

` 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.

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