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

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#

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#

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#