Find the gradient of the curve x=y+1/y at the point (2.5, 2). ?

3 Answers
Jul 23, 2018

The gradient is 4/3 at (2.5, 2)

Explanation:

Remember that the gradient of a tangent line at any point of the curve is just dy/dx.

First, differentiate x = y + 1/y using implicit differentiation.

color(red)(d/dx)x = color(red)(d/dx)(y + 1/y)

1 = (1 - 1/y^2)dy/dx

Notice that because y + 1/y was a function of x, by the chain rule, we must multiply by the derivative of y.

Now, isolate dy/dx.

dy/dx = 1/(1 - 1/y^2)

Now, we substitute color(red)(x = 2.5) and color(red)(y = 2).

dy/dx = 1/(1 - 1/color(red)(2)^2)

dy/dx = 4/3

4/3

Explanation:

Given function:

x=y+1/y

differentiating above equation w.r.t. x as follows

d/dx(x)=dy/dx+d/dx(1/y)

1=dy/dx-1/y^2 dy/dx

dy/dx(\frac{y^2-1}{y^2})=1

dy/dx=\frac{y^2}{y^2-1}

hence the gradient of the given curve dy/dx at the point (2.5, 2) is given by substituting y=2 in above differential equation as follows

dy/dx=\frac{2^2}{2^2-1}

=4/3

Jul 23, 2018

Please see the explanation below

Explanation:

Another method involving partial derivatives

Let

f(x,y)=x-y-1/y

The partial derivatives are

(delf)/(delx)=1

(delf)/(dely)=-1+1/y^2

Therefore,

dy/dx=-((delf)/(delx))/((delf)/(dely))=-1/(1/y^2-1)

At the point (2.5,2)

dy/dx=-1/(1/2^2-1)=1/(3/4)=4/3