*Mathematica* Q&A: Three Functions for Computing Derivatives

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This week’s question comes from Bashir, a student:

**What are the different functions for computing derivatives in Mathematica?**

The main function for computing derivatives in *Mathematica* is `D`, which computes the familiar partial derivative of an expression with respect to a variable:

`D` supports generalizations including multiple derivatives and derivatives with respect to multiple variables, such as differentiating twice with respect to *x*, then once with respect to *y*:

And vector and tensor derivatives, such as the gradient:

There are two important properties of `D`[*expr*, *x*] that distinguish it from other functions for computing derivatives in *Mathematica*:

1. `D` computes the derivative of an expression representing a quantity, such as `Sin``[ x]`, not a

*function*, such as

`Sin`. Compare with

`Derivative`below.

2. `D` takes the partial derivative to be zero for all subexpressions that don’t explicitly depend on *x*. Compare with `Dt` below.

To differentiate a function, you use `Derivative`, which has the standard shorthand notation `'` (apostrophe):

The result is a pure `Function` of one unnamed argument `#1`. Note that if you immediately evaluate this function at *x*, the result is exactly what you would have found by using `D` to differentiate the *quantity* `Sqrt`[*x*] with respect to *x*:

The notation f`'` is shorthand for `Derivative`[1][f], specifying differentiation once with respect to the first argument. As with `D`, generalizations like multiple derivatives and derivatives with respect to multiple variables are possible:

`Derivative`[2, 1] specifies differentiation twice with respect to the first argument and once with respect to the second argument. (You could also write the above in a single line as `Derivative[2, 1][#1^2 #2^3&]`.)

`Dt`[*expr*, *x*] computes the total derivative of the expression *expr* with respect to *x*. It works like `D`[*expr*, *e*], except `Dt` does not assume zero derivative for parts of *expr* with no dependence on *x*. Compare `D` and `Dt` in this short example:

`D` assumes *a* is a constant independent of *x*; `Dt` does not, and `Dt`[*a*, *x*] remains unevaluated.

This can be useful in situations where you have variables that implicitly depend on some other variable. For example, suppose you want the time derivative of x + y z given in terms of the time derivatives of *x*, *y*, and *z*. You can use `Dt`:

To do this with `D`, you would explicitly make *x*, *y*, and *z* functions of *t*:

Finally, the one-argument form `Dt`[*expr*] gives the total differential of the expression *expr*:

The result is given in terms of the differentials (`Dt`[*x*], `Dt`[*y*], in this case) of the variables occurring in *expr*.

When viewed in traditional mathematical notation (`TraditionalForm`), this example looks familiar as the standard quotient rule for differential quantities:

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What is sigma and how important is it?

Do I have to know it just for lean manufactorting or can I use a simple form?

Hello,

I have been unable to reconstruct the vey first example of this QA.

Thank you,

-peter