derivative

As you know, the Gradient of a function is the following vector: and the Hessian is the following matrix: Now, I wonder, is there any way to calculate these in R for a user defined function at a given point? First, I've found a package named numDeriv , which seems to have the necessary functions grad and hessian but now I can't get the correct results... Thus, here's my workflow: Let's say that we are given the function f(x,y) = x^2 * x^3, and we need to calculate the Gradient and the Hessian at the point (x=1, y=2). That's been said, I define this function within R: dummy <- function(x,y) {

I cannot seem to understand this, shouldn't the derivative/change along the U or V coordinate in a 2d texture/array be single dimension variable as we are checking it only along ddx (U coordinate) or ddy (V coordinate)? There are 4 distinct partial derivatives here: du/dx, dv/dx, du/dy, and dv/dy. None of those four values need be zero, unless the texture image coordinates happen to be perfectly aligned to the display screen axes. In general the texture coordinate axes need not be aligned to the screen display axes. X and Y (display viewport axes) are not the same directions as U and V

This question was migrated from Cross Validated because it can be answered on Stack Overflow. Migrated 2 years ago . Learn more . I'm trying to implement a function that computes the Relu derivative for each element in a matrix, and then return the result in a matrix. I'm using Python and Numpy. Based on other Cross Validation posts, the Relu derivative for x is 1 when x > 0, 0 when x < 0, undefined or 0 when x == 0 Currently, I have the following code so far: def reluDerivative(self, x): return np.array([self.reluDerivativeSingleElement(xi) for xi in x]) def reluDerivativeSingleElement(self,

For a neural networks library I implemented some activation functions and loss functions and their derivatives. They can be combined arbitrarily and the derivative at the output layers just becomes the product of the loss derivative and the activation derivative. However, I failed to implement the derivative of the Softmax activation function independently from any loss function. Due to the normalization i.e. the denominator in the equation, changing a single input activation changes all output activations and not just one. Here is my Softmax implementation where the derivative fails the

I wrote a simple derivative function in scheme today. I was asked to return a function such that g(x) = (f (x+h) -f(x))/h . Does this suffice to return a function or does this only return a value? (define (der f h) (lambda (x) (/ (- (f(+ x h)) (f x)) h))) Yes, the code in the question is returning a function (that's what the lambda is for). If it were returning a value , it'd be missing the line with (lambda (x) and the corresponding closing parentheses. Also notice that although the procedure is correct, the formula stated in the question isn't right, it should be: g(x) = (f(x+h) - f(x))/h ;