Autograd

Get PyTorch to compute a derivative for you.

For f(x) = 3x^2 + 2x, return df/dx at each value in x_values, using autograd rather than the derivative you can do in your head.

The mechanics:

x = torch.tensor(x_values, dtype=torch.float32, requires_grad=True)
y = 3 * x ** 2 + 2 * x
y.sum().backward()
x.grad

requires_grad=True is what makes torch record the operations. backward() walks that recording in reverse and accumulates the result into x.grad.

Calling backward() needs a scalar, because a gradient is only defined against one number. Summing first is the usual trick: each element of x affects exactly one element of y, so d(sum)/dx_i is dy_i/dx_i, which is what you wanted.

Return a tensor of shape (n,). You have already written this derivative by hand in the calculus problems; the point here is to see the same answer arrive without you deriving it.

Function signature

def grad_of_quadratic(x_values):

Example

Input

grad_of_quadratic([0.0, 1.0, 2.0])

Output

tensor([ 2.,  8., 14.])

Reasoning

df/dx = 6x + 2, which gives 2, 8 and 14 at 0, 1 and 2. Autograd finds the same values without you writing the derivative.

Constraints and notes

• Call backward() on a scalar, so sum y first.
• Return a tensor of shape (n,).