If f(x,y)=sin[x/(1+y)], calculate df/dx and df/dy.

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

f(x,y)= sin[x/(1+y)]

1. df/dx means we differentiate the function with respect to x while y is constant

then df/dx = [1/(1+y)]*cos(x/(1+y).

2. Now for df/dy we differentiate with respect to y while x is constant

==> df/dy = cos(x/(1+y)* [ -x/(1+y)^2]

              = -cos(x/(1+y)*[x/(1+y)^2

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

For calculating df/dx and df/dy, we'll use the chain rule, which we are usually using when differentiating functions of 1 variable.

To evaluate df/dx, we'll differentiate with respect to x, keeping y as constant.

df/dx ={cos[x/(1+y)]}*[1/(1+y)]

Now, to evaluate df/dy, we'll differentiate with respect to y, holding x as constant.

df/dy = cos{cos[x/(1+y)]}*[x'*(1+y)-x*(1+y)']/(1+y)^2

df/dy = cos{cos[x/(1+y)]}*[(0-x)/(1+y)^2]

df/dy = - cos{cos[x/(1+y)]}*[x/(1+y)^2]

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x,y) = sin(x/(x+y)). To calculate df/dx and df/dy.

Solution:

df/dx stands for the total differential coefficient with respect to x.

df/dx = partial derivative of f(x,y) with respect to x + partial derivative of f(x,y) with resppect to y* (dy/dx).

= [cos( x/(x+y))]*{1*(x+y) - x(1+0)}/(x+y)^2 + [cos (x/(x+y))]*(-x/(x+y)^2) * (dy/dx)

= [cos(x/(x+y))]{ (y/(x+y)^2 - [x/(x+y)^2] dy/dx }

df/dy  stands total differential coefficient of f with respect to y.

df/dy = partial differential coeeficient of f with respect to y +(partial diffrential coefficient of f with respect to x) * (dx/dy)

= [cos (x/(x+y))]* {-x/(x+y)^2} +[ cos (x/(x+y))] {[1*(x+y)-x*(1+0)]/(x+y)^2} (dx/dy)

=[cos(x/(x+y)] { -x/(x+y)^2  - [y/(x+y)^2] (dx/dy).

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