Question

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

Answer 1

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`

Answer 2

`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`

Answer 3

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`