Unit – 2 | unit 2 partial differentiation

Back to Study material

EM - 1

Unit – 2

Partial differentiation

Q 1:

Solution: 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

Q 2: 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

Q 3:

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)

Q 4: 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 :

$D \{ ( f(x) )^2 \} = 2 f(x) \ D \{ f(x) \} = 2 f(x) f'(x)$ .

Since y symbolically represents a function of x, the derivative of y2 can be found in the same fashion :

$D \{ y^2 \} = 2 y \ D \{ y \} = 2 y y'$ .

Now begin with

x2 + y2 = 25 .

Differentiate both sides of the equation, getting

D ( x2 + y2 ) = D ( 25 ) ,

D ( x2 ) + D ( y2 ) = D ( 25 ) ,

And

2x + 2 y y' = 0 ,

So that

2 y y' = - 2x ,

And

$y' = \displaystyle{ - 2x \over 2y } = \displaystyle{ - x \over y }$ ,

i.e.,

$y' = \displaystyle{ - x \over y }$ .

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

$m = y' = \displaystyle{ - (3) \over (-4) } = \displaystyle{ 3 \over 4 }$

Q 5: Find the total differential of w=x3yz + x y +z +3 at (1,2,3)

Solution: The total differential at the point (x0, y0, z0) is

Dw = wx (x0, y0, z0). Dx +wy( x0,y0,z0).dy + wz ( x0,y0,z0).dz

In our case,

wx=3x2yz+y wy=x3z+x wz=x3y+1

Substituting in the point (1,2,3) we get

wx(1,2,3) =20

wy(1,2,3) =4

wz(1,2,3) =3

Thus, the total differential is.,

Dw = 20dx+4dy+3dz.

Unit – 2

Partial differentiation

Q 1:

Solution: 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

Q 2: 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

Q 3:

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)

Q 4: 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:

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

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) .