Partial Derivatives
You have studied differentiation earlier and you might be thinking- how Partial Derivatives
is different from the regular differentiation? The difference between the two is itself the
definition of P.D.(the short form we'll be using for Partial Derivatives).
In Differentiation, we had two variables \(x, y\) where \(x\) was an independent variable and
\(y\) was dependent on \(x\), as shown in the diagram below:
Or in other words, a function having only one variable.
But what if we have more than one variable in a function? That's where P.D. comes in. In P.D. we find derivative of the function with respect to one of its variables, rest of the variables treated as constant, and repeat the same procedure with all of the variables. This means that all of the variables, unlike differentiation, are independent.
\[ \frac{\partial x}{\partial x} = 1, \frac{\partial y}{\partial x} = 0 \]
When partially differentiating w.r.t. \(x\), \(y\) is taken constant, hence its partial derivative \( = 0\).
Similarly,
\[ \frac{\partial y}{\partial y} = 1, \frac{\partial x}{\partial y} = 0 \]
Below are some examples that will clear the concept:
\[
\text{a)} \hspace{10pt}
\frac{\partial }{\partial x}(x^2 + y^2) = 2x + 0 ;\hspace{25pt}
\frac{\partial }{\partial y}(x^2 + y^2) = 0 + 2y
\]
\[
\text{b)} \hspace{10pt}
\frac{\partial }{\partial x}(xy) = y(1) = y ; \hspace{25pt}
\frac{\partial }{\partial y}(xy) = x(1) = x
\]
\[
\text{c)} \hspace{10pt}
\frac{\partial }{\partial x}(x^2y^2) = y^2(2x) = 2xy^2 ; \hspace{25pt}
\frac{\partial }{\partial y}(x^2y^2) = x^2(2y) = 2yx^2
\]
\[
\text{d)} \hspace{10pt}
\frac{\partial }{\partial x}(\sin{xy}) = \cos{xy}\times y(1) = y\cos{xy};
\]
\[
\frac{\partial }{\partial y}(\sin{xy}) = \cos{xy}\times x(1) = x\cos{xy}
\]
\[
\text{e)} \hspace{10pt}
\frac{\partial }{\partial x}(\log{x^2 + y^2}) = \frac{1}{x^2 + y^2}\times 2x = \frac{2x}{x^2 + y^2};
\]
\[
\frac{\partial }{\partial y}(\log{x^2 + y^2}) = \frac{1}{x^2 + y^2}\times 2y = \frac{2y}{x^2 + y^2}
\]
So, P.D. is quite simple, right? yeah, just take one variable at a time and the rest as constants. Actually, it's
not that simple, the process involved in differentiating can either be so simple that you can solve it without
lifting your pen or complicated enough to frustate you for not reaching to your answer, as we will see in sample
problems below. But before that, we need to know one more thing: identifying independent and dependent variables.
If there are 3 variables in the problem, 2 would be independent(mostly \(x\) and \(y\)) and one will be
dependent on the two (mostly \(z\)). Similarly, if there are 4 variables, 3 would be independent(\(x,y,z\)) and one
would be dependent on the three(\(u\)).
Sample Problem 1 :
Find first and second order partial drivatives of \[ z = x^3 + y^3 - 3x^2y^2 \]
Solution:
Simple process- differentiate w.r.t. one variable and treat rest as constant \[ \frac{\partial z}{\partial x} = 3x^2 + 0 - 6xy^2 = 3x^2 - 6xy^2 \] \[ \frac{\partial z}{\partial y} = 0 + 3y^2 - 6x^2y = 3y^2 - 6x^2y\] Same process for second order P.D. \[ \frac{\partial^2 z}{\partial x^2} = 6x - 6y^2 \] \[ \frac{\partial^2 z}{\partial y^2} = 6y - 6x^2 \]
Sample Problem 2 :
If \( z(x + y) = x^2 + y^2 \) then prove that \[ \left(\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} \right)^2 = 4\left(1 - \frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} \right) \]
Solution :
\[ z(x + y) = x^2 + y^2 \] \[ z = \left( \frac{x^2 + y^2}{x + y} \right) \hspace{25pt} \longrightarrow (1) \] Partially differentiate equation (1) w.r.t. x, \[ \frac{\partial z}{\partial x} = \frac{ (x + y)(2x) - (x^2 + y^2)(1) }{(x + y)^2} \] \[ = \frac{2x^2 + 2xy - x^2 - y^2}{(x + y)^2} \] \[ \frac{\partial z}{\partial x} = \frac{x^2 - y^2 + 2xy}{(x + y)^2} \] Now Partially differentiate equation (1) w.r.t. y, \[ \frac{\partial z}{\partial y} = \frac{(x + y)(2y) - (x^2 + y^2)(1)}{(x + y)^2} \] \[ \frac{\partial z}{\partial y} = \frac{y^2 - x^2 + 2xy}{(x + y)^2} \] L.H.s. , \[ \left(\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} \right)^2 = \left[\frac{x^2 - y^2 + 2xy - y^2 + x^2 - 2xy}{(x + y)^2} \right] \] \[ = \left[\frac{2(x^2 - y^2)}{(x + y)^2} \right]^2 \] \[ = \left[\frac{2\cancel{(x + y)}(x - y)}{(x + y)^\cancel 2} \right]^2 \] \[ = \frac{4(x - y)^2}{(x + y)^2} \] Now R.H.S., \[ 4\left(1 - \frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} \right) \] \[= 4\left[1 - \frac{x^2 - y^2 + 2xy}{(x + y)^2} - \frac{y^2 - x^2 + 2xy}{(x + y)^2} \right] \] \[= 4\left[\frac{\cancel{x^2} + y^2 + \cancel{2xy} - \cancel{x^2} + \cancel{y^2} - \cancel{2xy} - \cancel{y^2} + x^2 - 2xy}{(x + y)^2} \right] \] \[ = 4\left[\frac{x^2 + y^2 - 2xy}{(x + y)^2} \right] \] \[ = 4\frac{(x - y)^2}{(x + y)^2} \] = L.H.S., Hence Proved
Sample Problem 3 :
If \( u = log(x^3 + y^3 + z^3 - 3xyz) then show that \) \[ \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)^2u = \frac{-9}{(x + y + z)^2} \]
Solution :
It might look complicated but it's not. \[ u = log(x^3 + y^3 + z^3 - 3xyz) \hspace{25pt} \longrightarrow (1) \] L.H.S. = \[ \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)^2u \] \[ = \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)u \] \[ = \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)\left(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} \right) \hspace{20pt} \longrightarrow (2) \] Now partially differentiate equation (1) w.r.t. x \[ \frac{\partial u}{\partial x} = \frac{3x^2 - 3yz}{x^3 + y^3 + z^3 - 3xyz} \]Similarly, \[ \frac{\partial u}{\partial y} = \frac{3y^2 - 3xz}{x^3 + y^3 + z^3 - 3xyz} \] \[ \frac{\partial u}{\partial z} = \frac{3z^2 - 3xy}{x^3 + y^3 + z^3 - 3xyz} \] Putting the values in equation (2) \[ \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)\left[\frac{3x^2 - 3yz + 3y^2 - 3xz + 3z^2 - 3xy}{x^3 + y^3 + z^3 - 3xyz}\right] \] \[ = 3\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)\left[\frac{x^2 + y^2 + z^2 - yz - xz - xy}{x^3 + y^3 + z^3 - 3xyz}\right] \] \[ = 3\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)\left[\frac{\cancel{x^2 + y^2 + z^2 - yz - xz - xy}}{(x + y + z)\cancel{(x^2 + y^2 + z^2 - xy - yz - zx)}}\right] \] \[ \left[\text{As, } a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \right] \] \[ = 3\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)\left(\frac{1}{x + y + z} \right) \] \[ = 3\left[\frac{\partial }{\partial x}\left(\frac{1}{x + y + z} \right) + \frac{\partial }{\partial y}\left(\frac{1}{x + y + z} \right) + \frac{\partial }{\partial z}\left(\frac{1}{x + y + z} \right) \right] \] \[ = 3\left[-1(x + y + z)^{-2}(1) -1(x + y + z)^{-2}(1) -1(x + y + z)^{-2}(1) \right] \] \[ = \frac{-9}{(x + y + z)^2} \] = R.H.S., Hence Proved
