# If xy-5=x^3/2 determine dy/dx

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find dy/dx.

Now we are given that : xy - 5 = x^(3/2)

Differentiate both the sides with respect to x

=> d/dx (xy - 5) = d/dx( x^(3/2))

=> d/dx (xy) - d/dx(5) = d/dx( x^(3/2))

=> x*(dy/dx) + y - 0 = 3/2 * x^(1/2)

=> x*(dy/dx) = 3/2 * x^(1/2) - y

=> dy/dx = (3/2)*(1/x)* x^(1/2) - y/x

=> dy/dx = (3/2)*(x^-1/2) - y/x

Therefore dy/dx = (3/2)*(x^-1/2) - y/x.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

xy-5 = x^3/2.

To find dx/dy.

We xy -5 = x^3^2.

We diferentiate both sides with respect to x.

1y+xdy/dx = (3/2)x^(3/2-1).

y+xdy/dx = (3/2)x^(1/2).

xdy/dx = (3/2)x^(1/2)-y.

Therefore dy/dx = {(3/2)x^(1/2) -y}/x.

We take the reciprocal of both sides:

dx/dy = x/{(3/2)x^(1/2) -y}.

dx/dy = 2x/{3x^(1/2) -2y}.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To differentiate x wtih respect to y, dy/dx, we'll have to find a function of x.

xy-5=x^3/2

xy = x^3/2 + 5

y = [x^(3/2) + 5]/x

Now, we'll differentiate both sides:

dy/dx = {(d/dx)[x^(3/2) + 5]*x - [x^(3/2) + 5]*(d/dx)(x)}/x^2

dy/dx = [(3/2)*xsqrtx - xsqrtx - 5]/x^2

We'll combine like terms from numerator:

dy/dx = (xsqrtx - 10)/2x^2

dy/dx = sqrtx/2x - (5/x^2)