Question

Find f if grad f = (1/x)+(1/y)+(1/z).

Answer 1

Is it more like: \( \displaystyle \nabla f = \frac{1}{x} \hat{i} + \frac{1}{y} \hat{j} + \frac{1}{z} \hat{k} \)
With those components? I think if that's how it was written, the definition of gradient could be used with equating the like-components.
\( \displaystyle\nabla f = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k} \)

Answer 2

yup, which looks like it should be ln(x)+ln(y)+ln(z)+c, but webassign begs to differ...

Answer 3

well, the partial derivatives seem to bring up an interesting case, where you'd actually lose little function pieces of the other variables rather than a constant in partial differentiation...
Like if f is a function of x, y, and z, you'd lose any of the pieces without an x if you take that partial derivative w.r.t. x.

Answer 4

...Although, I am having trouble thinking of what makes your answer incorrect, since that solution does seem to work out, and I can't think of any way to find another working solution.

I am wondering if what you're answering this in has a specific handling on the constant you add in, if that's the standard indication of an additive constant or not. That could 'potentially' be a reason...

Answer 5

\[
\frac{\partial f}{\partial x} = \frac{1}{x} \\
\frac{\partial f}{\partial y} = \frac{1}{y} \\
\frac{ \partial f}{\partial z} = \frac{1}{z} \\
~\\
f = \ln x + g(y, z) \\
\frac{\partial f}{\partial y} = \frac{1}{y} \\
\frac{\partial f}{\partial z} = \frac{1}{z} \\
~\\
\frac{\partial}{\partial y} \left( \ln x + g(y, z) \right) = \frac{1}{y} \\
\frac{\partial}{\partial z} \left( \ln x + g(y, z) \right) = \frac{1}{z} \\
~\\
\frac{\partial g}{\partial y} = \frac{1}{y} \\
\frac{\partial g}{\partial z} = \frac{1}{z} \\
~\\
g = \ln y + h(z) \\
\frac{\partial}{\partial z} \left(\ln y + h(z) \right) = \frac{1}{z} \\
\frac{\partial h}{\partial z} = \frac{1}{z} \\
h = \ln z + c \\
g = \ln y + \ln z + c \\
f = \ln x + \ln y + \ln z + c \]
Yeah, I get the same answer here...