mattmns
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Just wondering if I did this right:
Here is the question: find \frac{\partial z}{\partial x} of \frac{x^2}{9} - \frac{y^2}{4} + \frac{z^2}{2} = 1
Now I put the \frac{\partial z}{\partial x} on both sides then got.
\frac{2x}{9} - 0 + z \frac{\partial z}{\partial x} = 0
So
\frac{\partial z}{\partial x} = -\frac{2x}{9z}
Now I know this is the right answer I am just curious if I did it right, it has been a while.
Now my reasoning:
\frac{\partial z}{\partial x} \frac{x^2}{9} is just the derivative with respect to x, so it will be \frac{2x}{9}
\frac{\partial z}{\partial x} \frac{y^2}{4} has no z or x, so it is constand and therefore 0.
\frac{\partial z}{\partial x} \frac{z^2}{2} has a z, so it is the derivative of itself, but times \frac{\partial z}{\partial x}
Is all of that correct, or did I do something wrong. The book I have only shows one example
Thanks!
Here is the question: find \frac{\partial z}{\partial x} of \frac{x^2}{9} - \frac{y^2}{4} + \frac{z^2}{2} = 1
Now I put the \frac{\partial z}{\partial x} on both sides then got.
\frac{2x}{9} - 0 + z \frac{\partial z}{\partial x} = 0
So
\frac{\partial z}{\partial x} = -\frac{2x}{9z}
Now I know this is the right answer I am just curious if I did it right, it has been a while.
Now my reasoning:
\frac{\partial z}{\partial x} \frac{x^2}{9} is just the derivative with respect to x, so it will be \frac{2x}{9}
\frac{\partial z}{\partial x} \frac{y^2}{4} has no z or x, so it is constand and therefore 0.
\frac{\partial z}{\partial x} \frac{z^2}{2} has a z, so it is the derivative of itself, but times \frac{\partial z}{\partial x}
Is all of that correct, or did I do something wrong. The book I have only shows one example

Thanks!
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