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Calculating second derivatives implicitly

by Moogie
Tags: derivatives, implicitly
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Moogie
#1
Jun11-10, 09:03 AM
P: 168
Hi

I've just been learning about how to get first derivatives implicitly and I think I'm getting it. Then the book comes onto calculating second derivatives implicitly and I don't know how to handle the dy/dx terms you might have in your equation from the first implicit differentiation

Lets say you have this after your first round of implicit differentiation

[tex]2\frac{dy}{dx} + cos(y)\frac{dy}{dx} = \frac{2x}{\Pi}[/tex]

Then lets get the second derivative of the first term

[tex]\frac{dy}{dx}\left(2 \frac{dy}{dx}\right)[/tex]

It seems so simple but I don't know how to do it. I know what the answer should be as its in the book. In fact the book thinks its so simple it doesn't bother to explain this term. But for some reason it's throwing me.

The derivative of dy/dx wrt to x is the second derivative d2y/dx^2 but I don't know how to take into account the 2. Is the 2 just classed as a multiple of a function which is why the answer is 2 (d2y/dx2)

thanks
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Mark44
#2
Jun11-10, 10:22 AM
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P: 21,215
Quote Quote by Moogie View Post
Hi

I've just been learning about how to get first derivatives implicitly and I think I'm getting it. Then the book comes onto calculating second derivatives implicitly and I don't know how to handle the dy/dx terms you might have in your equation from the first implicit differentiation

Lets say you have this after your first round of implicit differentiation

[tex]2\frac{dy}{dx} + cos(y)\frac{dy}{dx} = \frac{2x}{\Pi}[/tex]
Solving for dy/dx gives
[tex]\frac{dy}{dx}(2 + cos(y)) = \frac{2x}{\pi}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{2x}{\pi (2 + cos(y))}[/tex]

Now take the derivative with respect to x of both sides.
Quote Quote by Moogie View Post

Then lets get the second derivative of the first term

[tex]\frac{dy}{dx}\left(2 \frac{dy}{dx}\right)[/tex]
This doesn't mean what I think you intended.
This is dy/dx times 2dy/dx, which is 2(dy/dx)2. What you meant to write, I believe was
[tex]\frac{d}{dx}\left(2 \frac{dy}{dx}\right)[/tex]

This means "take the derivative of 2 times dy/dx."

dy/dx is the derivative of y with respect to x. d/dx is the operator that means to take the derivative of whatever quantity it's applied to. The first is a derivative; the second indicates intent.
Quote Quote by Moogie View Post

It seems so simple but I don't know how to do it. I know what the answer should be as its in the book. In fact the book thinks its so simple it doesn't bother to explain this term. But for some reason it's throwing me.

The derivative of dy/dx wrt to x is the second derivative d2y/dx^2 but I don't know how to take into account the 2. Is the 2 just classed as a multiple of a function which is why the answer is 2 (d2y/dx2)

thanks
HallsofIvy
#3
Jun11-10, 10:42 AM
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PF Gold
P: 39,345
Since "2" is a constant, the derivative of 2f(x), with respect to x, is 2 df/dx. In particular, the derivative of 2dy/dx is [itex]2d^2y/dx^2[/itex].

But, like Mark44, I suspect that is NOT the question you meant to ask! What does the problem say, really?

Moogie
#4
Jun11-10, 10:48 AM
P: 168
Calculating second derivatives implicitly

Hi

You are right about my typing error - thanks

You have rearranged the equation to make dy/dx the subject to get teh second derivative. That's not the way the book I am reading is doing it but that doesn't matter as it wasn't the actual problem itself I couldnt do - it was one step of the problem whcih confused me which was this bit in isolation

[tex]\frac{d}{dx}\left(2 \frac{dy}{dx}\right)[/tex]

I wondered what rule you are applying to get the answer which is 2(d^2d/dx^2). Is it just the constant multiple of a function rule? In other words the derivative of 2x^2 wrt x is simply 2 times the derivative of x^2 wrt x. Similarly the derivative of 2(dy/dx) wrt to x is 2 times the derivative of dy/dx wrt x.

I know this is simple but it confused me at the time and I wanted to check my understanding.

I think HallsOfIvy has answered at the same time I'm typing!
Moogie
#5
Jun11-10, 10:55 AM
P: 168
The actual problem was to get the second derivative of

[tex]2y + sin(y) = \frac{x^2}{pi} + 1[/tex]

It was a worked example and I could follow it except it confused me for some reasons on the term I pointed out.
phyzguy
#6
Jun11-10, 12:18 PM
P: 2,179
If you want to go through it nitpicking detail, you can use the product rule:
[tex]\frac{d}{dx}(2\frac{dy}{dx}) =\frac{d}{dx}(2)\frac{dy}{dx} +2 \frac{d}{dx}(\frac{dy}{dx}) [/tex]

Since 2 is a constant, it's derivative with respect to x is zero, so the first term is zero. So you are left with:

[tex]\frac{d}{dx}(2\frac{dy}{dx}) =2 \frac{d}{dx}(\frac{dy}{dx}) = 2 \frac{d^2y}{dx^2} [/tex]
Moogie
#7
Jun11-10, 01:24 PM
P: 168
That was why I confused myself! I wanted to use the product rule but I knew I didnt need to and got muddled. Thank-you!


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