I got question below:
import torch
x = torch.randn(size=(3,6), requires_grad=True)
t = torch.randn(size=(3,6), requires_grad=True)
y = 2 * torch.dot(x,t)
y.backward()
x.grad
t.grad
I try to create a function of two variable x and t, then do y.backward, but why I got error:
1D tensors expected, but got 2D and 2D tensors
torch.dot() is a function for vector multiply vector, use torch.mm() for matrix multiply matrix
Thanks, it works
Extra question
import torch
x = torch.randn(size=(3,6), requires_grad=True)
t = torch.randn(size=(6,4), requires_grad=True)
y = 2 * torch.mm(x,t)
y.sum().backward()
x.grad
t.grad
tensor([[ 3.3685, 3.3685, 3.3685, 3.3685],
[-4.0740, -4.0740, -4.0740, -4.0740],
[ 5.9460, 5.9460, 5.9460, 5.9460],
[ 0.3694, 0.3694, 0.3694, 0.3694],
[-3.9745, -3.9745, -3.9745, -3.9745],
[-2.2524, -2.2524, -2.2524, -2.2524]])
请问在2.5.4. Python控制流的梯度计算中
a = torch.randn(size=(), requires_grad=True)里面size()是什么意思呢？
size ( int… ) – a sequence of integers defining the shape of the output tensor. Can be a variable number of arguments or a collection like a list or tuple.
感谢回答，就是如果他是（3，4）我能明白他是想生成一个三行四列的矩阵，但是size=()不太明白是什么意思
size=()里面没有数是生成标量，有一个数就是向量，两个就是矩阵。
Thank you very much!!
import torch
import matplotlib.pyplot as plt
import numpy as np
x = torch.linspace(0, 3*np.pi, 128)
x.requires_grad_(True)
y = torch.sin(x) # y = sin(x)
y.sum().backward()
plt.plot(x.detach(), y.detach(), label='y=sin(x)')
plt.plot(x.detach(), x.grad, label='∂y/∂x=cos(x)') # dy/dx = cos(x)
plt.legend(loc='upper right')
plt.show()
Hi,
You got this error because the var x and y you created is 2D, and backward() expect a scalar input.
可以运行的，不过这里 post 的引号貌似都是中文的，需要自己改成英文引号
首先假设x是一个向量。f(x)对x求偏导数后，每个xi（向量x的分量）都会有自己的偏导数，所以x.grad和x的shape是相同的。如果x是矩阵结果也是一样的。
my solution for Question 5:
x = torch.linspace(-5, 5, 100)
x.requires_grad_(True)
y = torch.sin(x)
y.backward(torch.ones_like(x))
y = y.detach()
from utils import d2l
d2l.plot(x.detach(), [y, x.grad], 'f(x)', "f'(x)", legend=['f(x)', 'Tangent line'])
d2l.plt.show()
thanks. I run it successful
为什么按照PDF 上的 自动求导 y.backward().会报错。而且我也不理解为什么要求和再求导呢 y.sum().backward()
y 求和之后会是个数字。还怎么求梯度呢。不太理解什么意思。
import math
from d2l import torch as d2l
x=torch.arange(0,2*math.pi,0.01,requires_grad=True)
y=torch.sin(x)
y.backward(gradient=torch.ones(len(x)))
d2l.plot(x.detach(),[y.detach(),x.grad],‘x’,‘f(x)’,legend=[‘y’,‘df(x)/dx’])
请问样例代码中：
x = torch.arange(4.0)
x
tensor([0., 1., 2., 3.])
求导之后
x.grad
tensor([ 0., 4., 8., 12.])
为什么从14张量求导后还是14，而不是4*1呢？ 
for Q5:
# using plot function defined in 2.4
x = torch.arange(-2*torch.pi, torch.pi*2, 0.1,requires_grad=True)
f=torch.sin(x)
f.sum().backward()
f1=x.grad
x.requires_grad_(False)
plot(x, [torch.sin(x), f1], 'x', 'f(x)', legend=['f(x)=sin(x)', "f'(x)=cos(x)"])
关于自动求梯度函数 backward 的使用，这里有一个不错的讲解。