Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the difference between a gradient and a derivative? The text I'm reading keeps mentioning that 'the gradient is the transpose of the derivative'.

Does this mean that $$ \nabla f(v) = df(v)^T $$ and also that $$ \nabla f(v) = \frac{df(v)}{dx} $$

share|cite|improve this question
You can think of them as matrices of partial derivatives. The difference between the gradient and the derivative is how you arrange the partials in the matrix: one is the transpose of the other. – Isaac Solomon Mar 12 '12 at 3:22
So is $\nabla f(v) = \frac{df(v)}{dx}$ the correct definition of the gradient? As I understand it, the partials of the numerator are laid out according to the shape of f(v), while the denominator is laid out according to the transpose of $x$. – Chris D Mar 12 '12 at 5:30

The derivative of a function $f:\ {\mathbb R}^n\to{\mathbb R}^m$ at a point $p\in{\rm dom}(f)$ is a linear map $df(p):\ T_p \to T_q$ where $q:=f(p)$, $T_p\cong{\mathbb R}^n$ is the tangent space at $p$, and $T_q\cong{\mathbb R}^m$ is the tangent space at $q$.

As such $df(p)$ has a matrix with respect to the standard bases in $T_p$ and $T_q$, namely $$\bigl[df(p)\bigr]=\left[\matrix{{\partial f_1\over\partial x_1} & \cdots&{\partial f_1\over\partial x_n} \cr \vdots\cr {\partial f_m\over\partial x_1} & \cdots & {\partial f_m\over\partial x_n} \cr}\right]_p\ .$$

This makes sense for arbitrary $n\geq1$ and $m\geq1$. In the case at hand we have $m=1$, so the above matrix consists of one row only. Now in this case something special arises: The values of the "abstract" linear map $df(p):\ {\mathbb R}^n\to{\mathbb R}$ can be computed using the scalar product in ${\mathbb R}^n$. There is a "concrete" vector $a\in{\mathbb R}^n$ such for all vectors $X\in{\mathbb R}^n$ we have

$$df(p).X\ =\ a\cdot X\ \qquad(X\in T_p).\qquad(1)$$

This vector $a$ is nothing else but the gradient vector of $f$ at $p$, i.e., $$a =\nabla f(p):=\bigl(f_{.1}(p),\ldots, f_{.n}(p)\bigr)\ .$$ It is easy to see that for this $a$ the identity $(1)$ holds. In terms of matrices we can say the following: The components of $a$ are the entries in the single row of $\bigl[df(p)\bigr]$, and when you write $a$ (as is usual) as a column vector then this column vector, regarded as a matrix, is the transpose of the matrix $\bigl[df(p)\bigr]$.

share|cite|improve this answer

Generally, the derivative of a function $f:\mathbb R^n\to \mathbb R$ at a point $x$ is regarded as a linear map $\mathrm{d}f_x:\mathbb R^n\to \mathbb R$, while the gradient $\nabla f(x)$ if regarded as a vector in $\mathbb R^n$. The two objects (linear maps and vectors) are actually pretty interchangeable in finitely many dimensions, as the space of linear maps from $\mathbb R^n$ to $\mathbb R$, denoted $BL(\mathbb R^n,\mathbb R)$, is isomorphic to $\mathbb R^n$. The elements of $BL(\mathbb R^n,\mathbb R)$ are usually written as matrices with one row (aka column vectors). To move from $BL(\mathbb R^n,\mathbb R)$ to $\mathbb R^n$, we take the transpose. This maps $\mathrm{d}f_x$ to $\nabla f(x)$. Taking the transpose again gets us back to $\mathrm{d}f_x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.