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

Question

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

3 Answers

Impact of this question

9520 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-23T12:16:49

The gradient is $\frac{4}{3}$ $4/3$ at $(2.5, 2)$ $(2.5, 2)$

Explanation:

Remember that the gradient of a tangent line at any point of the curve is just $\frac{d y}{d x}$ $dy/dx$ .

First, differentiate $x = y + \frac{1}{y}$ $x = y + 1/y$ using implicit differentiation.

$\frac{d}{d x} x = \frac{d}{d x} (y + \frac{1}{y})$ $color(red)(d/dx)x = color(red)(d/dx)(y + 1/y)$

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

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

Now, isolate $\frac{d y}{d x}$ $dy/dx$ .

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

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

$\frac{d y}{d x} = \frac{1}{1 - \frac{1}{2^{2}}}$ $dy/dx = 1/(1 - 1/color(red)(2)^2)$

$\frac{d y}{d x} = \frac{4}{3}$ $dy/dx = 4/3$

Answer 2 · 2018-07-23T12:22:46

$\frac{4}{3}$ $4/3$

Explanation:

Given function:

$x = y + \frac{1}{y}$ $x=y+1/y$

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

$\frac{d}{d x} (x) = \frac{d y}{d x} + \frac{d}{d x} (\frac{1}{y})$ $d/dx(x)=dy/dx+d/dx(1/y)$

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

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

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

hence the gradient of the given curve $\frac{d y}{d x}$ $dy/dx$ at the point $(2.5, 2)$ $(2.5, 2)$ is given by substituting $y = 2$ $y=2$ in above differential equation as follows

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

$= \frac{4}{3}$ $=4/3$

Answer 3 · 2018-07-23T12:23:19

Please see the explanation below

Explanation:

Another method involving partial derivatives

Let

$f (x, y) = x - y - \frac{1}{y}$ $f(x,y)=x-y-1/y$

The partial derivatives are

$\frac{\partial f}{\partial x} = 1$ $(delf)/(delx)=1$

$\frac{\partial f}{\partial y} = - 1 + \frac{1}{y^{2}}$ $(delf)/(dely)=-1+1/y^2$

Therefore,

$\frac{d y}{d x} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = - \frac{1}{\frac{1}{y^{2}} - 1}$ $dy/dx=-((delf)/(delx))/((delf)/(dely))=-1/(1/y^2-1)$

At the point $(2.5, 2)$ $(2.5,2)$

$\frac{d y}{d x} = - \frac{1}{\frac{1}{2^{2}} - 1} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$ $dy/dx=-1/(1/2^2-1)=1/(3/4)=4/3$

Science

Math

Humanities

... and beyond

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

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question