Implicit differentiation

Inbox

Hi I'm tackling some new problems. Can somebody please advise me.

Q. Check that the given values for x and y satisfy the given equation, then use implicit differentiation to calculate dy/dx for these values.

x=-2 and y=1 for the equation 2y^2-1/y+3x+5 =0

So I'm wondering do I put the x and y value in at the end? and also what do I do to the 1/y term, what does that change into? Thanks

sullanefc

Inbox wrote: »

Hi I'm tackling some new problems. Can somebody please advise me.

Q. Check that the given values for x and y satisfy the given equation, then use implicit differentiation to calculate dy/dx for these values.

x=-2 and y=1 for the equation 2y^2-1/y+3x+5 =0

So I'm wondering do I put the x and y value in at the end? and also what do I do to the 1/y term, what does that change into? Thanks

The question is asking you to put in the x and y values in to the equation above to check that it is on the curve first.

Then differentiate, and put in the x and y values again. (This is usually used to find the slope of the tangent to the curve at that point).

On the 1/y term, write it as y^-1 and then differentiate as you normally would for implicit differentiation.

Inbox

Thank you

joeperry

Hi i have one here that i'm finding tricky.

Use implicit differentiation too calculate dy/dx

2xy-ln(3y)=2x^3+y^2

The answer i got is,

6x^2-2y / 2x-1/3y-2y


The ln(3y) is what im unsure of, i'm thinking it becomes 1/3y? Thanks

TheBody

joeperry wrote: »

Hi i have one here that i'm finding tricky.

Use implicit differentiation too calculate dy/dx

2xy-ln(3y)=2x^3+y^2

The answer i got is,

6x^2-2y / 2x-1/3y-2y


The ln(3y) is what im unsure of, i'm thinking it becomes 1/3y? Thanks

Hi Joe,

Yea, you have a small error in the derivative of the [latex]\ln(3y)[/latex] bit.

If you think about this:
[latex]\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}[/latex].

So in your case:
[latex]\frac{d}{dx}\ln(3y)=\frac{3}{3y}\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}[/latex]

Hope that makes sense!!

joeperry

Thanks, i still not totally sure but i'll battle on. Hey how are you typing out the maths bit. What program is that? Thanks.

joeperry

Sorry just a quick one if i had ln(2-3x) does that become 1/2-3x? Wolframalpha is telling me it's 3/3x-2.

TheBody

joeperry wrote: »

Sorry just a quick one if i had ln(2-3x) does that become 1/2-3x? Wolframalpha is telling me it's 3/3x-2.

[latex]\frac{d}{dx}\ln(2-3x)=\frac{-3}{2-3x}[/latex] (which is the same thing as Wolframalpha with a little manipulation).

The derivative of the log of a function is the derivative of the function over the original function. It's easy enough to prove using the chain rule.

TheBody

joeperry wrote: »

Thanks, i still not totally sure but i'll battle on. Hey how are you typing out the maths bit. What program is that? Thanks.

I'm typing the maths stuff using a typesetting code called Latex. There is a whole forum on it here:

http://www.boards.ie/vbulletin/forumdisplay.php?f=1286

It is amazingly powerful for writing maths stuff but it does take some effort to learn it.

joeperry

OK thanks i'm getting it now