- #1
Moogie
- 168
- 1
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 let's 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
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 let's 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