Mathematics College

If F(x,y) = x^2sin(xy), find Fyx.

Answers

Answer 1

Answer:

[tex]F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)[/tex]

Step-by-step explanation:

We need to find out the partial differential [tex]F_{yx}[/tex] of [tex]F(x,y)=x^{2}sin(xy)[/tex]

First, differentiate [tex]F(x,y)=x^{2}sin(xy)[/tex] both the sides with respect to 'y'

[tex]\frac{d}{dy}F(x,y)=\frac{d}{dy}x^{2}sin(xy)[/tex]

Since, [tex]\frac{d}{dt}\sin t =\cos t[/tex]

[tex]\frac{d}{dy}F(x,y)=x^{2}cos(xy)\times \frac{d}{dy}(xy)[/tex]

[tex]\frac{d}{dy}F(x,y)=x^{2}cos(xy)\times x[/tex]

[tex]\frac{d}{dy}F(x,y)=x^{3}cos(xy)[/tex]

so, [tex]F_y=x^{3}cos(xy)[/tex]

Now, differentiate above both the sides with respect to 'x'

[tex]F_{yx}=\frac{d}{dx}x^{3}cos(xy)[/tex]

Chain rule of differentiation: [tex]D(fg)=f'g + fg'[/tex]

[tex]F_{yx}=cos(xy) \frac{d}{dx}x^{3} + x^{3} \frac{d}{dx}cos(xy)[/tex]

Since, [tex] \frac{d}{dx}x^{m} =mx^{m-1}[/tex] and [tex] \frac{d}{dt} cost =-\sin t[/tex]

[tex]F_{yx}=cos(xy)\times 3x^{2} - x^{3} sin(xy)\times \frac{d}{dx}(xy)[/tex]

[tex]F_{yx}=cos(xy)\times 3x^{2} - x^{3} sin(xy)\times y[/tex]

[tex]F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)[/tex]

hence, [tex]F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)[/tex]

Related Questions

Find the area under the curve y =f( x) on [a,b] given f(x)=tan(3x) where a=0 b=pi/12

Answers

Answer:

The area under the curve y=f(x) on [a,b] is [tex]\frac{1}{6}\ln(2)[/tex] square units.

Step-by-step explanation:

The given function is

[tex]f(x)=\tan(3x)[/tex]

where a=0 and b=pi/12.

The area under the curve y=f(x) on [a,b] is defined as

[tex]Area=\int_{a}^{b}f(x)dx[/tex]

[tex]Area=\int_{0}^{\frac{\pi}{12}}\tan (3x)dx[/tex]

[tex]Area=\int_{0}^{\frac{\pi}{12}}\frac{\sin (3x)}{\cos (3x)}dx[/tex]

Substitute cos (3x)=t, so

[tex]-3\sin (3x)dx=dt[/tex]

[tex]\sin (3x)dx=-\frac{1}{3}dt[/tex]

[tex]a=\cos (3(0))=1[/tex]

[tex]b=\cos (3(\frac{\pi}{12}))=\frac{1}{\sqrt{2}}[/tex]

[tex]Area=-\frac{1}{3}\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{t}dt[/tex]

[tex]Area=-\frac{1}{3}[\ln t]_{1}^{\frac{1}{\sqrt{2}}[/tex]

[tex]Area=-\frac{1}{3}(\ln \frac{1}{\sqrt{2}}-\ln (1))[/tex]

[tex]Area=-\frac{1}{3}(\ln 1-\ln \sqrt{2}-0)[/tex]

[tex]Area=-\frac{1}{3}(-\ln 2^{\frac{1}{2}})[/tex]

[tex]Area=-\frac{1}{3}(-\frac{1}{2}\ln 2)[/tex]

[tex]Area=-\frac{1}{6}\ln 2[/tex]

Therefore the area under the curve y=f(x) on [a,b] is [tex]\frac{1}{6}\ln(2)[/tex] square units.

(05.05 MC) The area of a triangle is 24 square inches. What is the height of the triangle if the base length is 8 inches?

Answers

height: 6

formula: 1/2bh

hope this helps :)

For this case we have that by definition, the area of a triangle is given by:

[tex]A = \frac {1} {2} b * h[/tex]

Where:

b: It's the base

h: It's the height

They tell us as data that:

[tex]A = 24 \ in ^ 2\\b = 8in[/tex]

Substituting the data and clearing the height:

[tex]24 = \frac {1} {2} 8 * h\\24 = 4h\\h = \frac {24} {4}\\h = 6[/tex]

So, the height of the triangle is 6in

Answer:

[tex]h = 6in[/tex]

you begin playing a new game called hooville. You are King of Hooville, a city of owls that is located in the treetops near Fords of Beruna. In order to know how much food to produce each year, you must predict the population of Hooville. History shows that the population growth rate of Hooville is 3.5%. The current population of owls is 80,000. Using the monetary growth formula that you used in the Uncle Harold problem, write a new function for the population of hooville. (let n=1.) PLEASE HELP. I HAVE NO IDEA WHAT IM DOING!!

Answers

Answer:

Part 1) [tex]y=80,000(1.035)^{x}[/tex]

Part 2) The table in the attached figure

Part 3) The graph in the attached figure

Step-by-step explanation:

Part 1) Find the population function

In this problem we have a exponential function of the form

[tex]y=a(b)^{x}[/tex]

where

y ----> is the population

x ----> the time in years

a is the initial value (a=80,000 people)

b is the base (b=100%+3.5%=103.5%=1.035)

substitute

[tex]y=80,000(1.035)^{x}[/tex]

Part 2) Construct the table

For x=0 years