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DrClaude said:In ##L_x##, it should be ##\cos \phi##, not ##\cot \phi##. ##L_z## has the wrong sign.
Not sure what you mean here.rasi said:So how can i go on. In no way i coulnd't tackled it. Thanks for your help...
DrClaude said:Not sure what you mean here.
Can you describe the problem you want to solve? The title of the thread is not very clear, do you mean you need to write ##L_x^2 + L_y^2 + L_z^2## in spherical coordinates ##(\theta, \phi)##?
The derivative of Lx^2+Ly^2+Lz^2 is 2Lx+2Ly+2Lz. This can be found by using the power rule for derivatives, which states that for any term of the form ax^n, the derivative is nx^(n-1).
To find the derivative of Lx^2+Ly^2+Lz^2, you can use the power rule for derivatives. This rule states that for any term of the form ax^n, the derivative is nx^(n-1). In this case, we have three terms, each with a coefficient of L and an exponent of 2, so the derivative will be 2Lx+2Ly+2Lz.
Yes, the derivative of Lx^2+Ly^2+Lz^2 can be simplified. In fact, the derivative we found in the previous questions, 2Lx+2Ly+2Lz, is already in its simplest form. However, if we were to factor out the common factor of 2L, we could rewrite the derivative as 2L(x+y+z).
No, the derivative of Lx^2+Ly^2+Lz^2 is not constant. The derivative is a function that represents the rate of change of the original function. In this case, the derivative is equal to 2Lx+2Ly+2Lz, which means that the rate of change is different for each variable (x, y, and z) and is dependent on the value of L.
The derivative of Lx^2+Ly^2+Lz^2 is related to the original function in that it represents the rate of change of the original function. This means that the derivative can give us information about how the values of x, y, and z are changing with respect to the value of L. Additionally, the derivative can also help us find the slope of the tangent line to the original function at a specific point.