Partial Derivatives

Partial derivatives, as the title implies, involves differentiating a function with respect to a particular variable assuming all the others are kept constant, this concept is called partial differentiation, it involves differentiating a function partially with respect to a particular variable though the function contains several variables, consider $u$ which is a function of $x$ and $y$, represented as $u(x,y)$:

$u= x^2+y^2$

taking the partial derivative of $u$ with respect to $x$:

$\frac{\partial \, u}{\partial x}=$

$\frac{\partial \, (x^2+y^2)}{\partial x}=$

$\frac{\partial \, (x^2)}{\partial x}+\frac{\partial \, (y^2)}{\partial x}=$

$2x+0=$

$2x$

Note from the above partial differentiation, that the partial derivative of a function of a single variable with respect to that variable is the same as the ordinary derivative of that function:

$\frac{\partial \, (x^2)}{\partial x}= \frac{d \, (x^2)}{dx}=2x $

because $x^2$ is a function of $x$ denoted by f(x).

Also note that the partial derivative of a function of a single variable with respect to a different variable is $0$, which is as taking the ordinary derivative of any constant with respect to any variable:

$\frac{\partial \, (y^2)}{\partial x}= \frac{d \, (9)}{dx} = \frac{d \, (k)}{dz}=0$

assuming $k$ is a constant.

Example:

Take the partial derivatives of the following functions with respect to y:

(a) $2x^3+2y^5+1$

(b) $5x+6y+z^2$

(c) $zx+x^3 $

Example Solutions:

(a)

$\frac{ \partial \, (2x^3+2y^5+1) }{\partial y}=$

$\frac{\partial \, (2x^3)}{\partial y}+\frac{\partial \, (2y^5)}{\partial y}+\frac{\partial \, (1)}{\partial y}=$

$10y^4$

(b)

$\frac{\partial \, (5x+6y+z^2)}{\partial y}=$

$\frac{\partial \, (5x)}{\partial y}+\frac{\partial \, (6y)}{\partial y}+\frac{\partial \, (z^2)}{\partial y}=$

$6$

(c)

$\frac{\partial \, (zx+x^3)}{\partial y}=$

$\frac{\partial \, (zx)}{\partial y}+\frac{\partial \, (x^3)}{\partial y}=$

$0$

Final Inference:

The partial derivative of a constant k with respect to any variable is $0$:

$\frac{\partial \, (k)}{\partial z}=0$

Excercise:

Simplify the following:

(a) $\frac{\partial \, (x+y+z)}{\partial z}$

(b) $\frac{\partial \, (2t^5+5y)}{\partial t}$

(c) $\frac{\partial \, (\frac{xy-i^2}{2})}{\partial x}$

Excercise Solutions:

(a)

$\frac{\partial \, (x+y+z)}{\partial z}=1$

(b)

$\frac{\partial \, (2t^5+5y)}{\partial t}=10t^4$

(c)

$\frac{\partial \, (\frac{xy-i^2}{2})}{\partial x}=\frac{y}{2}$