Unit 2 | unit 2 partial differentiation

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EM - I

Unit - 2

Partial differentiation

2.1 Partial derivative and geometrical interpretation

Partial Differentiation of composite function

a) Let and , then z becomes a function of , In this case z is called composite function of .

i.e.

b) Let possess continuous partial derivatives and let possess continuous partial derivatives, then z is called composite function of x and y.

i.e.

Continuing in this way, …..

Ex. If Then prove that

Ex. If then prove that

Where is function of x, y, z.

Ex. If where ,

then show that,

Ii)

Notations of partial derivatives of variable to be treated as a constant

Let

and

i.e.

Then means partial derivative of u w.r.t. x treating y const.

To find from given reactions we first express x in terms of u & v.

i.e. & then diff. x w.r.t. u treating v constant.

To find express v as a function of y and u i.e. then diff. v w.r.t. y treating u as a const.

Ex. If , then find the value of

Ex. If , then prove that

2.2 Euler’s theorem with corollaries and their applications

A polynomial in x & y is said to be Homogeneous expression in x & y of degree n. If the degree of each term in the expression is same & equal to n.

e.g.

is a homogeneous function of degree 3.

To find the degree of homogeneous expression f(x, y).

Consider
Put . Then if we get .

Then the degree of is n.

Ex.

Consider

Put

Thus degree of f(x, y) is

Note that

If be a homogeneous function of degree n then z can be written as

Differentiation of Implicit function

Suppose that we cannot find y explicitly as a function of x. But only implicitly through the relation f(x, y) = 0.

Then we find

Since

diff. P. w.r.t. x we get

i.e.

Similarly,

It f (x, y, z) = 0 then z is called implicit function of x, y. Then in this case we get

Ex.

Find if

Ex. Find . If , &

Ex. If , where

Find

Ex. If

Then find

Eulers Theorem on Homogeneous functions:

Statement:

If be a homogeneous function of degree n in x & y then,

Deductions from Eulers theorem

If be a homogeneous function of degree n in x & y then,

2. If be a homogeneous functions of degree n in x & y and also then,

And

Where

Ex.

If , find the value of

Ex.

If then find the value of

Ex. If then prove

That

Ex. If the prove that

Ex. If then show

That

2.3 Chain rule

In calculus, the general Leibniz rulenamed after Gottfried Wilhelm Leibniz, generalizes the product rule (which is also known as "Leibniz's rule"). It states that if {\displaystyle f}f and {\displaystyle g}g are {\displaystyle n}n-times differentiable functions, then the product {\displaystyle fg}fg is also {\displaystyle n}n-times differentiable and its {\displaystyle n}nth derivative is given by

(fg)n = k=0n (nk) f(n-k).g(k)

{\displaystyle (f\circ g)'=(f'\circ g)\cdot g'.}Alternatively, by letting $F$ $=$ $f$ $\circ$ $g$ (equiv., $F$ $($ $x$ $) =$ $f$ $($ $g$ $($ $x$ $))$ for all $x$ ), one can also write the chain rule in Lagrange's notation, as follows:

F’(X)= f’(g(x)).g’(x)

{\displaystyle F'(x)=f'(g(x))g'(x).}The chain rule may also be rewritten in Leibniz's notation in the following way. If a variable $z$ depends on the variable $y$ , which itself depends on the variable $x$ (i.e., $y$ and $z$ are dependent variables), then $z$ , via the intermediate variable of $y$ , depends on $x$ as well. In which case, the chain rule states that:

{\displaystyle {\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}.} ${\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}.$ =

The Chain Rule Formula is as follows –

= .

Solved Examples

Example 1: Differentiate y = cos x2

Solution:

Given,

y = cos x2

Let u = x2, so that y = cos u

Therefore: =2x

= -sin u

And so, the chain rule says:

= .

= -sin u × 2x

= -2x sin x2

Example 2:

Differentiate f(x)=(1+x2)5.

Solution:

Using the Chain rule,

Let us take y = u5 and u = 1+x2

Then = (u5) = 5u4

= (1 + x2 )= 2x.

= 5u4⋅2x = 5(1+x2)4⋅2x

= 10x(1+x4)

3. {\displaystyle (fg)^{(n)}=\sum _{k=0}^{n}{n \choose k}f^{(n-k)}g^{( Implicit differentiation

Implicit differentiation:

A function can be explicit or implicit:

Explicit: "y = some function of x". When we know x we can calculate y directly.

Implicit: "some function of y and x equals something else". Knowing x does not lead directly to y.

How to do Implicit Differentiation

Differentiate with respect to x
Collect all the on one side
Solve for

Example 1: x2 + y2 = r2

Differentiate with respect to x:

(x2) + (y2) = (r2)

Let's solve each term:

Use the Power Rule: (x2) = 2x

Use the Chain Rule (explained below): (y2) = 2y

r2 is a constant, so its derivative is 0:d/dx(r2) = 0

Which gives us:

2x + 2y= 0

Collect all the on one side

Y = −x

Solve for :

The Chain Rule Using :

Let's look more closely at how (y2) becomes 2y

The Chain Rule says:

Substitute in u = y2:

(y2) = (y2) .

And then:

(y2) = 2y

Basically, all we did was differentiate with respect to y and multiply by dy/dx

Example 2:

The following problems require the use of implicit differentiation. Implicit differentiation is nothing more than a special case of the well-known chain rule for derivatives. The majority of differentiation problems in first-year calculus involve functions y written EXPLICITLY as functions of x. For example, if

$y = 3x^2 -\sin(7x+5)$ ,

Then the derivative of y is

$y' = 6x - 7 \cos(7x+5)$ .

However, some functions y is written IMPLICITLY as functions of x. A familiar example of this is the equation

x2 + y2 = 25,

Which represents a circle of radius five cantered at the origin. Suppose that we wish to find the slope of the line tangent to the graph of this equation at the point (3, -4) .

https://www.math.ucdavis.edu/~kouba/IMPLICITgraphsdirectory/Implicit.gif

How could we find the derivative of y in this instance? One way is to first write y explicitly as a function of x. Thus,

x2 + y2 = 25,

y2 = 25 - x2,

And

$y = \pm \sqrt{ 25 - x^2 }$ ,

Where the positive square root represents the top semi-circle and the negative square root represents the bottom semi-circle. Since the point

(3, -4) lies on the bottom semi-circle given by

$y = - \sqrt{ 25 - x^2 }$ ,

The derivative of y is

$y' = - (1/2) \big( 25 - x^2 \big)^{-1/2} (-2x) = \displaystyle{ x \over \sqrt{ 25 - x^2 } }$ ,

i.e.,

$y' = \displaystyle{ x \over \sqrt{ 25 - x^2 } }$ .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

$m = y' = \displaystyle{ (3) \over \sqrt{ 25 - (3)^2 } } = \displaystyle{ 3 \over 4 }$ .

Unfortunately, not every equation involving x and y can be solved explicitly for y . For the sake of illustration we will find the derivative of y WITHOUT writing y explicitly as a function of x . Recall that the derivative (D) of a function of x squared, (f(x))2 , can be found using the chain rule :