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Implicit Differentiation - Logarithm

Implicit Differentiation Proper Answer?Implicit Differentiation ProblemImplicit differentiationImplicit differentiation of $x2^y=ln y$Implicit Differentiation involving trigonometric functions.Difficulty Understanding Implicit DifferentiationImplicit Differentiation under conditionsHelp regarding implicit differentiation.Implicit differentiation to obtain expression valueQuotient rule and implicit differentiation

0

$begingroup$

$xlog(x) + ylog(y) = 1$

$dfracdydx= ?$

I calculated $fracdydx= -frac1+log(x)1+log(y)$

however, the correct answer seems to be $-log(x)/log(y)$

I'm confused, can someone help?

implicit-differentiation

share|cite|improve this question

edited Mar 25 at 15:59

Shubham Johri

5,475818

asked Mar 25 at 15:44

Aayush MalikAayush Malik

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE! Could you please explain how you achieved your solution?
  $endgroup$
  – Diglett
  Mar 25 at 15:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your solution is correct.
  $endgroup$
  – PierreCarre
  Mar 25 at 15:57
  

  
  
  
  

add a comment | 

0

$begingroup$

$xlog(x) + ylog(y) = 1$

$dfracdydx= ?$

I calculated $fracdydx= -frac1+log(x)1+log(y)$

however, the correct answer seems to be $-log(x)/log(y)$

I'm confused, can someone help?

implicit-differentiation

share|cite|improve this question

edited Mar 25 at 15:59

Shubham Johri

5,475818

asked Mar 25 at 15:44

Aayush MalikAayush Malik

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE! Could you please explain how you achieved your solution?
  $endgroup$
  – Diglett
  Mar 25 at 15:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your solution is correct.
  $endgroup$
  – PierreCarre
  Mar 25 at 15:57
  

  
  
  
  

add a comment | 

$begingroup$

$xlog(x) + ylog(y) = 1$

$dfracdydx= ?$

I calculated $fracdydx= -frac1+log(x)1+log(y)$

however, the correct answer seems to be $-log(x)/log(y)$

I'm confused, can someone help?

implicit-differentiation

share|cite|improve this question

edited Mar 25 at 15:59

Shubham Johri

5,475818

asked Mar 25 at 15:44

Aayush MalikAayush Malik

1

$endgroup$

share|cite|improve this question

edited Mar 25 at 15:59

Shubham Johri

5,475818

asked Mar 25 at 15:44

Aayush MalikAayush Malik

1

share|cite|improve this question

edited Mar 25 at 15:59

Shubham Johri

5,475818

asked Mar 25 at 15:44

Aayush MalikAayush Malik

1

edited Mar 25 at 15:59

Shubham Johri

5,475818

edited Mar 25 at 15:59

Shubham Johri

5,475818

asked Mar 25 at 15:44

Aayush MalikAayush Malik

1

asked Mar 25 at 15:44

Aayush MalikAayush Malik

1

2 Answers
2

active

oldest

votes

1

$begingroup$

I got $$log(x)+1+y'log(y)+y'=0$$

share|cite|improve this answer

answered Mar 25 at 15:47

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

0

$begingroup$

Assuming that $log(x)=log_e(x)=ln(x)$, we start off with our equation:

$$xlog(x)+ylog(y)=1$$

Next we perform implicit differentiation to both sides:

$$
fracdydx(xlog(x)+ylog(y))=fracdydx(1)\
1+log(x)+fracdydx1+fracdydxlog(y)=0
$$

Now we isolate $fracdydx$ to one side:

$$
1+log(x)+fracdydx1+fracdydxlog(y)=0\
fracdydx1+fracdydxlog(y)=-log(x)-1\
fracdydx(1+log(y))=-log(x)-1\
fracdydx=frac-log(x)-11+log(y)\
$$

Finally we clean things up and simplify:

$$beginalign
fracdydx&=frac-log(x)-11+log(y)\
&bbox[5px,border:2px solid black]
=-frac1+log(x)1+log(y)

endalign$$

We see that your solution is indeed correct.

share|cite|improve this answer

answered Mar 28 at 22:09

CizoxCizox

266

$endgroup$

add a comment | 

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1

$begingroup$

I got $$log(x)+1+y'log(y)+y'=0$$

share|cite|improve this answer

answered Mar 25 at 15:47

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

1

$begingroup$

I got $$log(x)+1+y'log(y)+y'=0$$

share|cite|improve this answer

answered Mar 25 at 15:47

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

$begingroup$

I got $$log(x)+1+y'log(y)+y'=0$$

share|cite|improve this answer

answered Mar 25 at 15:47

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

share|cite|improve this answer

answered Mar 25 at 15:47

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

answered Mar 25 at 15:47

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

answered Mar 25 at 15:47

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

0

$begingroup$

Assuming that $log(x)=log_e(x)=ln(x)$, we start off with our equation:

$$xlog(x)+ylog(y)=1$$

Next we perform implicit differentiation to both sides:

$$
fracdydx(xlog(x)+ylog(y))=fracdydx(1)\
1+log(x)+fracdydx1+fracdydxlog(y)=0
$$

Now we isolate $fracdydx$ to one side:

$$
1+log(x)+fracdydx1+fracdydxlog(y)=0\
fracdydx1+fracdydxlog(y)=-log(x)-1\
fracdydx(1+log(y))=-log(x)-1\
fracdydx=frac-log(x)-11+log(y)\
$$

Finally we clean things up and simplify:

$$beginalign
fracdydx&=frac-log(x)-11+log(y)\
&bbox[5px,border:2px solid black]
=-frac1+log(x)1+log(y)

endalign$$

We see that your solution is indeed correct.

share|cite|improve this answer

answered Mar 28 at 22:09

CizoxCizox

266

$endgroup$

add a comment | 

0

$begingroup$

Assuming that $log(x)=log_e(x)=ln(x)$, we start off with our equation:

$$xlog(x)+ylog(y)=1$$

Next we perform implicit differentiation to both sides:

$$
fracdydx(xlog(x)+ylog(y))=fracdydx(1)\
1+log(x)+fracdydx1+fracdydxlog(y)=0
$$

Now we isolate $fracdydx$ to one side:

$$
1+log(x)+fracdydx1+fracdydxlog(y)=0\
fracdydx1+fracdydxlog(y)=-log(x)-1\
fracdydx(1+log(y))=-log(x)-1\
fracdydx=frac-log(x)-11+log(y)\
$$

Finally we clean things up and simplify:

$$beginalign
fracdydx&=frac-log(x)-11+log(y)\
&bbox[5px,border:2px solid black]
=-frac1+log(x)1+log(y)

endalign$$

We see that your solution is indeed correct.

share|cite|improve this answer

answered Mar 28 at 22:09

CizoxCizox

266

$endgroup$

add a comment | 

$begingroup$

Assuming that $log(x)=log_e(x)=ln(x)$, we start off with our equation:

$$xlog(x)+ylog(y)=1$$

Next we perform implicit differentiation to both sides:

$$
fracdydx(xlog(x)+ylog(y))=fracdydx(1)\
1+log(x)+fracdydx1+fracdydxlog(y)=0
$$

Now we isolate $fracdydx$ to one side:

$$
1+log(x)+fracdydx1+fracdydxlog(y)=0\
fracdydx1+fracdydxlog(y)=-log(x)-1\
fracdydx(1+log(y))=-log(x)-1\
fracdydx=frac-log(x)-11+log(y)\
$$

Finally we clean things up and simplify:

$$beginalign
fracdydx&=frac-log(x)-11+log(y)\
&bbox[5px,border:2px solid black]
=-frac1+log(x)1+log(y)

endalign$$

We see that your solution is indeed correct.

share|cite|improve this answer

answered Mar 28 at 22:09

CizoxCizox

266

$endgroup$

share|cite|improve this answer

answered Mar 28 at 22:09

CizoxCizox

266

answered Mar 28 at 22:09

CizoxCizox

266

answered Mar 28 at 22:09

CizoxCizox

266