Implicit Differentiation

[Silvanoshei]

Test.... tex works. Good.

Original problem. \(\displaystyle \sin \frac{x}{y}=\frac{1}{2}\)

My work so far.

\(\displaystyle \frac{d}{dx}\sin \frac{x}{y}\cdot \frac{y\frac{d}{dx}x-x\frac{d}{dx}y}{y^{2}}=0\)

\(\displaystyle \cos \frac{x}{y}\cdot \frac{y-x\frac{dy}{dx}}{y^{2}}=0\)

Now... here's the problem, I get stuck because i'm not seeing why the book says it goes to \(\displaystyle y-x\frac{d_{y}}{dx}=0\)?

I get \(\displaystyle \cos \frac{x}{y}\cdot \frac{-x\frac{dy}{dx}}{y}=0\)?

Edit note: how to make tex equations bigger? seems a little small to read.

Edit 2: Fixed size

 

[srmichael]

Test.... tex works. Good.

Original problem. \(\displaystyle \sin \frac{x}{y}=\frac{1}{2}\)

My work so far.

\(\displaystyle \frac{d}{dx}\sin \frac{x}{y}\cdot \frac{y\frac{d}{dx}x-x\frac{d}{dx}y}{y^{2}}=0\)

\(\displaystyle \cos \frac{x}{y}\cdot \frac{y-x\frac{dy}{dx}}{y^{2}}=0\)

Now... here's the problem, I get stuck because i'm not seeing why the book says it goes to \(\displaystyle y-x\frac{d_{y}}{dx}=0\)?

I get \(\displaystyle \cos \frac{x}{y}\cdot \frac{-x\frac{dy}{dx}}{y}=0\)?

Edit note: how to make tex equations bigger? seems a little small to read.

Edit 2: Fixed size

I'm not sure what you mean when you say "the book says it goes to \(\displaystyle y-x\frac{d_{y}}{dx}=0\)?

I can tell you that it looks like you canceled the y in the numerator with one of the y's in the denominator and you can not do that since the y in the numerator is not a separate factor.

 

[Silvanoshei]

My question is, how does the book go from:

\(\displaystyle \cos \frac{x}{y}\cdot \frac{y-x\frac{dy}{dx}}{y^{2}}=0\)

to:

\(\displaystyle y-x\frac{d_{y}}{dx}=0\)?

 

[srmichael]

My question is, how does the book go from:

\(\displaystyle \cos \frac{x}{y}\cdot \frac{y-x\frac{dy}{dx}}{y^{2}}=0\)

to:

\(\displaystyle y-x\frac{d_{y}}{dx}=0\)?

Because when a fraction equals zero, only the numerator can equal zero.

 

[DrPhil]

Test.... tex works. Good.

Original problem. \(\displaystyle \sin \frac{x}{y}=\frac{1}{2}\)

My work so far.

\(\displaystyle \frac{d}{dx}\sin \frac{x}{y}\cdot \frac{y\frac{d}{dx}x-x\frac{d}{dx}y}{y^{2}}=0\)

\(\displaystyle \cos \frac{x}{y}\cdot \frac{y-x\frac{dy}{dx}}{y^{2}}=0\) OK this far

Now... here's the problem, I get stuck because i'm not seeing why the book says it goes to \(\displaystyle y-x\frac{d_{y}}{dx}=0\)?

I get \(\displaystyle \cos \frac{x}{y}\cdot \frac{-x\frac{dy}{dx}}{y}=0\)?

Edit note: how to make tex equations bigger? seems a little small to read.

Edit 2: Fixed size

Separate the two terms, then solve for dy/dx

\(\displaystyle \dfrac{1}{y} \cos\left({\dfrac{x}{y}}\right) - \dfrac{x}{y^2} \cos\left({\dfrac{x}{y}}\right) \dfrac{dy}{dx} = 0\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{y}{x} \)

Note that you can divide out the cosine because it is not zero (actual value is sqrt(3)/2). It is also "safe" to divide by x because the original equation does not allow the sine to be 0.

 

[Silvanoshei]

Confused on that seperation of the two terms.

Are we just using that zero to cancel out the
\(\displaystyle \cos \frac{x}{y}+y^{2}\)?

 

[DrPhil]

I'll put in the intermediate steps.
Separate the two terms, then solve for dy/dx

\(\displaystyle \dfrac{1}{y} \cos\left({\dfrac{x}{y}}\right) - \dfrac{x}{y^2} \cos\left({\dfrac{x}{y}}\right) \dfrac{dy}{dx} = 0\)

\(\displaystyle \dfrac{x}{y^2} \cos\left({\dfrac{x}{y}}\right) \dfrac{dy}{dx} = \dfrac{1}{y} \cos\left({\dfrac{x}{y}}\right) \)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{1}{y} \cos\left({\dfrac{x}{y}}\right)}{\dfrac{x}{y^2} \cos\left({\dfrac{x}{y}}\right)}\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{y}{x} \)

Note that you can divide out the cosine because it is not zero (actual value is sqrt(3)/2). It is also "safe" to divide by x because the original equation does not allow the sine to be 0.[/QUOTE]