How to Find the Limit of a Function's Derivative and Graph It?

[aruwin]

I have to find the limit of a function and draw a graph of it. First, can you check my work and then tell me how to draw the graph?


When 2 variables x,y are related by y=1+xlogy, find lim(x->1) dy/dx.

So,my first step is differentiating both sides:
dy/dx = logy + x*(1/y)*(dy/dx)
dy/dx = (y*logy)/(y - x)

So, lim(x->1) dy/dx = lim(x->1) (y*logy)/(y - x) = y*logy/(y-1)

Did I get the limit right? So how to draw the graph?

 

[chisigma]

I have to find the limit of a function and draw a graph of it. First, can you check my work and then tell me how to draw the graph?


When 2 variables x,y are related by y=1+xlogy, find lim(x->1) dy/dx.

So,my first step is differentiating both sides:
dy/dx = logy + x*(1/y)*(dy/dx)
dy/dx = (y*logy)/(y - x)

So, lim(x->1) dy/dx = lim(x->1) (y*logy)/(y - x) = y*logy/(y-1)

Did I get the limit right? So how to draw the graph?

We have a function y=f(x) implicity defined as...

$\displaystyle g(x,y)= y -x\ \ln y -1 =0$ (1)

... and, according to the Dini's Theorem, its derivative is ...

$\displaystyle f^{\ '} (x)= - \frac{g^{\ '}_{x}(x.y)}{g^{\ '}_{y}(x,y)}= \frac{\ln y}{1-\frac{x}{y}}$ (2)

For x=1 the (1) has the solution y=1 so that [applying l'Hopital's rule...] is...

$\displaystyle \lim_{x \rightarrow 1} f^{\ '}(x) = \lim_{y \rightarrow 1} \frac{y\ \ln y}{y-1} = \lim_{y \rightarrow 1} (1+\ln y)=1$ (3)

Kind regards

$\chi$ $\sigma$

 

[aruwin]

We have a function y=f(x) implicity defined as...

$\displaystyle g(x,y)= y -x\ \ln y -1 =0$ (1)

... and, according to the Dini's Theorem, its derivative is ...

$\displaystyle f^{\ '} (x)= - \frac{g^{\ '}_{x}(x.y)}{g^{\ '}_{y}(x,y)}= \frac{\ln y}{1-\frac{x}{y}}$ (2)

For x=1 the (1) has the solution y=1 so that [applying l'Hopital's rule...] is...

$\displaystyle \lim_{x \rightarrow 1} f^{\ '}(x) = \lim_{y \rightarrow 1} \frac{y\ \ln y}{y-1} = \lim_{y \rightarrow 1} (1+\ln y)=1$ (3)

Kind regards

$\chi$ $\sigma$

Got it :) Now how to plot the graph? What sould I do first?

 

[chisigma]

Got it :) Now how to plot the graph? What sould I do first?

I think that the graph is requested to find f(1) and for that wolframalpha is very comfortable...

http://www.wolframalpha.com/input/?i=y+-+ln+y+-1=0

Kind regards

$\chi$ $\sigma$

 

[aruwin]

I think that the graph is requested to find f(1) and for that wolframalpha is very comfortable...

http://www.wolframalpha.com/input/?i=y+-+ln+y+-1=0

Kind regards

$\chi$ $\sigma$

Actually, you're right. The graph is requested to find f(1).

 

[HallsofIvy]

I have to find the limit of a function and draw a graph of it. First, can you check my work and then tell me how to draw the graph?


When 2 variables x,y are related by y=1+xlogy, find lim(x->1) dy/dx.

So,my first step is differentiating both sides:
dy/dx = logy + x*(1/y)*(dy/dx)
dy/dx = (y*logy)/(y - x)

So, lim(x->1) dy/dx = lim(x->1) (y*logy)/(y - x) = y*logy/(y-1)

Did I get the limit right? So how to draw the graph?

Well, the obviousd thing to do is to write the equation as x= \frac{y- 1}{log y}. Choose a number of different values of y, calculate the corresponding value of x, and plot (x, y). Once you get sufficient points, draw a smooth curve through them.
(You say to begin with that you "have to find the limit of a function" but then say "find lim(x->1) dy/dx". Are you to find the limit of the given function or its derivative?)

 

[aruwin]

Well, the obviousd thing to do is to write the equation as x= \frac{y- 1}{log y}. Choose a number of different values of y, calculate the corresponding value of x, and plot (x, y). Once you get sufficient points, draw a smooth curve through them.
(You say to begin with that you "have to find the limit of a function" but then say "find lim(x->1) dy/dx". Are you to find the limit of the given function or its derivative?)

Its derivative.