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Logan Land
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To solve the gradient f when f = ln |r| do I start with differentiating each x,y,z term of the vector?Like ln|x| ln|y|...etc.
View attachment 4146
View attachment 4146
i tried to follow how I attempted the next question, gradient f if f=1/rFantini said:Notice that $\ln(\| {\mathbf r} \|) = \ln (\sqrt{x^2+y^2+z^2})$. You need to use the chain rule for the gradient. I don't understand your notation. :(
Fantini said:Remember the gradient is $$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right).$$ Since we've established $$\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2+y^2+z^2}}$$ and likewise for the other partial derivatives, we have $$\nabla f = \frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}} = \frac{{\mathbf r}}{r} = \frac{\widehat{ {\mathbf r}}}{r^2}.$$ :) Hope this helps. What do you mean with 'zbar'?
Patricio Lima said:I did not understand. gradiente (1/r) = - r/ r^3 ??
The formula for solving gradient with ln is:∇f(x,y) = (∂f/∂x, ∂f/∂y) = (1/x, 1/y)
Yes, the chain rule can be used when solving gradient with ln. The derivative of ln(x) is 1/x, and the chain rule allows us to find the derivative of a composite function.
No, logarithmic differentiation is not necessary when solving gradient with ln. The derivative of ln(x) can be found using the chain rule and the properties of logarithms.
The natural logarithm, ln(x), is the inverse of the exponential function, e^x. The gradient represents the rate of change of a function at a given point, and the natural logarithm can be used to find the gradient of exponential functions.
One common mistake is forgetting to apply the chain rule when finding the derivative of a function involving ln(x). It is also important to remember the properties of logarithms, such as ln(ab) = ln(a) + ln(b) and ln(a/b) = ln(a) - ln(b).