In mathematics, sometimes the function depends on two or more than two variables. In this case, the derivative converts into the partial derivative since the function depends on several variables. Whereas, partial differential equation, is an equation containing one or more partial derivatives is called a partial differential equation. Partial derivatives are usually used in vector calculus and differential geometry. In this article students will learn the basics of partial differentiation.
Partial Derivative Rules Just like ordinary derivatives, partial derivatives follows some rule like product rule, quotient rule, chain rule etc.
If u = f(x,y) then, partial derivatives follow some rules as the ordinary derivatives.
Product Rule:
If u = f(x,y).g(x,y), then
u x u_{x} u x ∂ u ∂ x \frac{\partial u}{\partial x} ∂ x ∂ u g ( x , y ) g\left ( x,y \right ) g ( x , y ) ∂ f ∂ x \frac{\partial f}{\partial x} ∂ x ∂ f + f ( x , y ) + f\left ( x,y \right ) + f ( x , y ) ∂ g ∂ x \frac{\partial g}{\partial x} ∂ x ∂ g And, u y u_{y} u y ∂ u ∂ y \frac{\partial u}{\partial y} ∂ y ∂ u g ( x , y ) g\left ( x,y \right ) g ( x , y ) ∂ f ∂ y \frac{\partial f}{\partial y} ∂ y ∂ f + f ( x , y ) + f\left ( x,y \right ) + f ( x , y ) ∂ g ∂ y \frac{\partial g}{\partial y} ∂ y ∂ g
Quotient Rule:
If u = f ( x , y ) g ( x , y ) \frac{f(x,y)}{g(x,y)} g ( x , y ) f ( x , y ) ≠ \neq =
u x u_{x} u x g ( x , y ) ∂ f ∂ x − f ( x , y ) ∂ g ∂ x [ g ( x , y ) ] 2 \frac{g\left ( x,y \right )\frac{\partial f}{\partial x}-f\left ( x,y \right )\frac{\partial g}{\partial x}}{\left [ g\left ( x,y \right ) \right ]^{2}} [ g ( x , y ) ] 2 g ( x , y ) ∂ x ∂ f − f ( x , y ) ∂ x ∂ g And, u y u_{y} u y g ( x , y ) ∂ f ∂ y − f ( x , y ) ∂ g ∂ y [ g ( x , y ) ] 2 \frac{g\left ( x,y \right )\frac{\partial f}{\partial y}-f\left ( x,y \right )\frac{\partial g}{\partial y}}{\left [ g\left ( x,y \right ) \right ]^{2}} [ g ( x , y ) ] 2 g ( x , y ) ∂ y ∂ f − f ( x , y ) ∂ y ∂ g
Power Rule:
If u = [f(x,y)]2 then, partial derivative of u with respect to x and y defined as
u x = n [ f ( x , y ) ] n – 1 u_{x} = n\left [ f\left ( x,y \right ) \right ]^{n – 1} u x = n [ f ( x , y ) ] n – 1 ∂ f ∂ x \frac{\partial f}{\partial x} ∂ x ∂ f And, u y = n [ f ( x , y ) ] n – 1 u_{y} = n\left [ f\left ( x,y \right ) \right ]^{n – 1} u y = n [ f ( x , y ) ] n – 1 ∂ f ∂ y \frac{\partial f}{\partial y} ∂ y ∂ f
Chain Rule:
If u = f(x,y) is a function where, x = (s,t) and y = (s,t) then by the chain rule, we can find the partial derivatives us and ut as:
u s u_{s} u s ∂ u ∂ x . ∂ x ∂ s + ∂ u ∂ y . ∂ y ∂ s \frac{\partial u}{\partial x}.\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial s} ∂ x ∂ u . ∂ s ∂ x + ∂ y ∂ u . ∂ s ∂ y and u t u_{t} u t ∂ u ∂ x . ∂ x ∂ t + ∂ u ∂ y . ∂ y ∂ t \frac{\partial u}{\partial x}.\frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial t} ∂ x ∂ u . ∂ t ∂ x + ∂ y ∂ u . ∂ t ∂ y
Partial Derivative Examples Given below are some of the examples on Partial Derivatives.
Question 1: Determine the partial derivative of a function fx and fy : if f(x, y) is given by f(x, y) = tan(xy) + sin x
Solution:
Given function is f(x, y) = tan(xy) + sin x
Derivative of a function with respect to x is given as follows:
fx = ∂ f ∂ x \frac{\partial f}{\partial x} ∂ x ∂ f ∂ ∂ x \frac{\partial}{\partial x} ∂ x ∂ [ tan ( x y ) + sin x ] [\tan (xy) + \sin x] [ tan ( x y ) + sin x ]
= ∂ ∂ x \frac{\partial}{\partial x} ∂ x ∂ [ tan ( x y ) ] + [\tan(xy)] + [ tan ( x y ) ] + ∂ ∂ x \frac{\partial}{\partial x} ∂ x ∂ [ sin x ] [\sin x] [ sin x ]
= sec2 (xy). y + cos x
= y sec2 (xy) + cos x
Now, Derivative of a function with respect to y. So, x is constant
fy = ∂ f ∂ y \frac{\partial f}{\partial y} ∂ y ∂ f ∂ ∂ y \frac{\partial}{\partial y} ∂ y ∂ [ tan ( x y ) + sin x ] [\tan (xy) + \sin x] [ tan ( x y ) + sin x ]
= ∂ ∂ y \frac{\partial}{\partial y} ∂ y ∂ [ tan ( x y ) ] + [\tan (xy)] + [ tan ( x y ) ] + ∂ ∂ y \frac{\partial}{\partial y} ∂ y ∂ [ sin x ] [\sin x] [ sin x ]
= sec2 (xy) x + 0
= x sec2 (xy)
Answer: fx = y sec2 (xy) + cos x and fy = x sec2 (xy)
Question 2: If f(x,y) = 2x + 3y, where x = t and y = t2 . Derivative f with respect to t.
Solution:
We know, d f d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} d t d f = ∂ x ∂ f d t d x + ∂ y ∂ f d t d y
Then, ∂ f ∂ x \frac{\partial f}{\partial x} ∂ x ∂ f ∂ f ∂ y \frac{\partial f}{\partial y} ∂ y ∂ f d x d t \frac{dx}{dt} d t d x d y d t \frac{dy}{dt} d t d y
So, we get
d f d t \frac{df}{dt} d t d f
Question 3: If f = x 2 ( y – z ) + y 2 ( z – x ) + z 2 ( x – y ) f = x^{2}(y – z) + y^{2}(z – x) + z^{2}(x – y) f = x 2 ( y – z ) + y 2 ( z – x ) + z 2 ( x – y ) ∂ f ∂ x \frac {\partial f} {\partial x} ∂ x ∂ f ∂ f ∂ y \frac {\partial f} {\partial y} ∂ y ∂ f ∂ f ∂ z \frac {\partial f} {\partial z} ∂ z ∂ f + 0 + 0 + 0
Solution:
Given, f = x 2 ( y – z ) + y 2 ( z – x ) + z 2 ( x – y ) f = x^{2} (y – z) + y^{2}(z – x) + z^{2}(x – y) f = x 2 ( y – z ) + y 2 ( z – x ) + z 2 ( x – y )
To find ∂ f ∂ x \frac {\partial f} {\partial x} ∂ x ∂ f
∂ f ∂ x \frac {\partial f} {\partial x} ∂ x ∂ f 2 x ( y – z ) + 0 ( 0 – 1 ) + 0 ( 1 – 0 ) 2x(y – z) + 0(0 – 1) + 0(1 – 0) 2 x ( y – z ) + 0 ( 0 – 1 ) + 0 ( 1 – 0 ) ∂ f ∂ x \frac {\partial f} {\partial x} ∂ x ∂ f 2 x y – 2 x z + 0 + 0 2xy – 2xz + 0 + 0 2 x y – 2 x z + 0 + 0 ∂ f ∂ x \frac {\partial f} {\partial x} ∂ x ∂ f 2 x y – 2 x z 2xy – 2xz 2 x y – 2 x z
To find ∂ f ∂ y \frac {\partial f} {\partial y} ∂ y ∂ f
∂ f ∂ y \frac {\partial f} {\partial y} ∂ y ∂ f 0 ( 1 – z ) + 2 y ( z – x ) + 0 ( 0 – 1 ) 0(1 – z) + 2y(z – x) + 0(0 – 1) 0 ( 1 – z ) + 2 y ( z – x ) + 0 ( 0 – 1 ) ∂ f ∂ y \frac {\partial f} {\partial y} ∂ y ∂ f 0 + 2 y z – 2 y x + 0 0 + 2yz – 2yx + 0 0 + 2 y z – 2 y x + 0 ∂ f ∂ y \frac {\partial f} {\partial y} ∂ y ∂ f 2 y z – 2 y x 2yz – 2yx 2 y z – 2 y x
To find ∂ f ∂ z \frac {\partial f} {\partial z} ∂ z ∂ f
∂ f ∂ z \frac {\partial f} {\partial z} ∂ z ∂ f 0 ( 0 – 1 ) + 0 ( 1 – 1 ) + 2 z ( x – y ) 0(0 – 1) + 0(1 – 1) + 2z(x – y) 0 ( 0 – 1 ) + 0 ( 1 – 1 ) + 2 z ( x – y ) ∂ f ∂ z \frac {\partial f} {\partial z} ∂ z ∂ f 0 + 0 + 2 z x – 2 z y 0 + 0 + 2zx – 2zy 0 + 0 + 2 z x – 2 z y ∂ f ∂ z \frac {\partial f} {\partial z} ∂ z ∂ f 2 z x – 2 z y 2zx – 2zy 2 z x – 2 z y
Adding (i) , (ii) and (iii) we get
∂ f ∂ x \frac {\partial f} {\partial x} ∂ x ∂ f ∂ f ∂ y \frac {\partial f} {\partial y} ∂ y ∂ f ∂ f ∂ z \frac {\partial f} {\partial z} ∂ z ∂ f 2 x y – 2 x z + 2 y z – 2 y x – 2 z x – 2 z y 2xy – 2xz + 2yz – 2yx – 2zx – 2zy 2 x y – 2 x z + 2 y z – 2 y x – 2 z x – 2 z y ∂ f ∂ x \frac {\partial f} {\partial x} ∂ x ∂ f ∂ f ∂ y \frac {\partial f} {\partial y} ∂ y ∂ f ∂ f ∂ z \frac {\partial f} {\partial z} ∂ z ∂ f
Hence proved.
Question 4: Given F = sin (xy ). Show that ∂2 F / (∂x ∂y) is equal to ∂2 F / (∂y ∂x).
Solution:
F = sin ( x y )
∂ F / ∂ x = y cos ( x y )
∂ F / ∂ y = x cos ( x y )
∂^ 2 F / ∂ x ∂ y = ∂ / ∂ y [ y cos ( x y ) ]
∂^ 2 F / ∂ x ∂ y = − x y sin ( x y ) + cos ( x y )
∂^ 2 F / ∂ y ∂ x = ∂ / ∂ x [ x cos ( x y ) ]
∂^ 2 F / ∂ y ∂ x = − x y sin ( x y ) + cos ( x y )
Hence,
∂2 F / (∂x ∂y) =∂2 F / (∂y ∂x)
Question 5: f (x, y) = x^2 + xy + y^2 , x = uv, y = u/v. To show that ufu + vfv = 2xfx and ufu − vfv = 2yfy
Solution: We need to find fu, fv, fx and fy.
fu = ∂f / ∂u = [∂f/ ∂x] [∂x / ∂u] + [∂f / ∂y] [∂y / ∂u];
fv = ∂f / ∂v = [∂f / ∂x] [∂x / ∂v] + [∂f / ∂y] [∂y / ∂v].
fu = (2x + y)(v) + (x + 2y)(1 / v) = 2uv^2 + 2u + 2u / v^2 .
fv = (2x + y)(u) + (x + 2y)(−u / v^2 ) = 2u^2 v − 2u^2 / v^3 .
fx = 2x + y = 2uv + u / v .
So, 2xfx = 2uvfx = 4u^2v^2 + 2u^2 .
fy = x + 2y = uv + 2u / v .
So, 2yfy = [2u / v] fx = 2u^2 + 4u^2 / v^2 .
Now ufu + vfv = 2u^2 v^2 + 2u^2 + 2u^2 / v^2 + 2u^2 v^2 − 2u^2 / v^2
= 4u^2 v^2 + 2u^2 = 2xfx as required,
and ufu − vfv = 2u^2 v^2 + 2u^2 + 2u^2 / v^2 − 2u^2 v^2 + 2u^2 / v^2
= 2u^2 + 4u^2 / v^2 = 2yfy as required.
Question 6: Show that the largest triangle of the given perimeter is equilateral.
Solution:
A = √
Where
P = given
s = [1 / 2] P
a+b+c=P
c=P–a–b ← equation (1)
A = 1 2 P ( 1 2 P – a ) ( 1 2 P – b ) [ 1 2 P – ( P – a – b ) ] A = 1 2 P ( 1 2 P – a ) ( 1 2 P – b ) ( a + b – 1 2 P ) ∂ A ∂ a = 1 2 P ( 1 2 P – b ) [ ( 1 2 P – a ) ( 1 ) + ( a + b – 1 2 P ) ( − 1 ) ] 2 1 2 P ( 1 2 P – a ) ( 1 2 P – b ) ( a + b – 1 2 P ) = 0 1 2 P – a = a + b – 1 2 P A = \sqrt{\frac{1}{2}P(\frac{1}{2}P – a)(\frac{1}{2}P – b)[\frac{1}{2}P – (P – a – b)]}\\ A = \sqrt{\frac{1}{2}P(\frac{1}{2}P – a)(\frac{1}{2}P – b)(a + b – \frac{1}{2}P)}\\ \frac{\partial A}{\partial a} = \frac{\frac{1}{2}P(\frac{1}{2}P – b)\left[ (\frac{1}{2}P – a)(1) + (a + b – \frac{1}{2}P)(-1) \right]}{2\sqrt{\frac{1}{2}P(\frac{1}{2}P – a)(\frac{1}{2}P – b)(a + b – \frac{1}{2}P)}} = 0\\\frac{1}{2}P – a = a + b – \frac{1}{2}P\\ A = 2 1 P ( 2 1 P – a ) ( 2 1 P – b ) [ 2 1 P – ( P – a – b ) ] A = 2 1 P ( 2 1 P – a ) ( 2 1 P – b ) ( a + b – 2 1 P ) ∂ a ∂ A = 2 2 1 P ( 2 1 P – a ) ( 2 1 P – b ) ( a + b – 2 1 P ) 2 1 P ( 2 1 P – b ) [ ( 2 1 P – a ) ( 1 ) + ( a + b – 2 1 P ) ( − 1 ) ] = 0 2 1 P – a = a + b – 2 1 P b = P – 2a ← equation (2)
∂ A ∂ b = 1 2 P ( 1 2 P – a ) [ ( 1 2 P – b ) ( 1 ) + ( a + b – 1 2 P ) ( − 1 ) ] 2 1 2 P ( 1 2 P – a ) ( 1 2 P – b ) ( a + b – 1 2 P ) = 0 1 2 P – b = a + b – 1 2 P \frac{\partial A}{\partial b} = \frac{\frac{1}{2}P(\frac{1}{2}P – a)\left[ (\frac{1}{2}P – b)(1) + (a + b – \frac{1}{2}P)(-1) \right]}{2\sqrt{\frac{1}{2}P(\frac{1}{2}P – a)(\frac{1}{2}P – b)(a + b – \frac{1}{2}P)}} = 0\\\frac{1}{2}P – b = a + b – \frac{1}{2}P\\ ∂ b ∂ A = 2 2 1 P ( 2 1 P – a ) ( 2 1 P – b ) ( a + b – 2 1 P ) 2 1 P ( 2 1 P – a ) [ ( 2 1 P – b ) ( 1 ) + ( a + b – 2 1 P ) ( − 1 ) ] = 0 2 1 P – b = a + b – 2 1 P a = P – 2b
Substitute b = P – 2a of equation (2)
a = P – 2 ( P – 2 a ) a = P – 2 P + 4 a 3 a = P a = 1 3 P a = P – 2(P – 2a)\\ a = P – 2P + 4a\\ 3a = P\\ a = \frac{1}{3}P\\ a = P – 2 ( P – 2 a ) a = P – 2 P + 4 a 3 a = P a = 3 1 P From equation (2)
b = P – 2 ( 1 3 P ) b = 1 3 P b = P – 2(\frac{1}{3}P)\\ b = \frac{1}{3}P\\ b = P – 2 ( 3 1 P ) b = 3 1 P From equation (1)
c = P – 1 3 P – 1 3 P c = 1 3 P H e n c e , a = b = c = P / 3. c = P – \frac{1}{3}P – \frac{1}{3}P\\ c = \frac{1}{3}P\\ \text \ Hence, \ a \ = \ b \ = \ c \ = \ P/3. c = P – 3 1 P – 3 1 P c = 3 1 P H e n c e , a = b = c = P / 3 . T h u s , t h e t r i a n g l e i s e q u i l a t e r a l . T h e m a x i m u m a r e a o f t h e t r i a n g l e i s : A m a x = 1 2 P ( 1 2 P – 1 3 P ) ( 1 2 P – 1 3 P ) ( 1 2 P – 1 3 P ) A m a x = 1 2 P ( 1 6 P ) ( 1 6 P ) ( 1 6 P ) A m a x = 1 432 P 4 A m a x = 1 12 3 P 2 \text \ Thus, \ the \ triangle \ is \ equilateral.\\ \text \ The \ maximum \ area \ of \ the \ triangle \ is\ \ : A_{max} = \sqrt{\frac{1}{2}P(\frac{1}{2}P – \frac{1}{3}P)(\frac{1}{2}P – \frac{1}{3}P)(\frac{1}{2}P – \frac{1}{3}P)}\\ A_{max} = \sqrt{\frac{1}{2}P(\frac{1}{6}P)(\frac{1}{6}P)(\frac{1}{6}P)}\\ A_{max} = \sqrt{\frac{1}{432}P^4}\\ A_{max} = \frac{1}{12\sqrt{3}}P^2\\ T h u s , t h e t r i a n g l e i s e q u i l a t e r a l . T h e m a x i m u m a r e a o f t h e t r i a n g l e i s : A m a x = 2 1 P ( 2 1 P – 3 1 P ) ( 2 1 P – 3 1 P ) ( 2 1 P – 3 1 P ) A m a x = 2 1 P ( 6 1 P ) ( 6 1 P ) ( 6 1 P ) A m a x = 4 3 2 1 P 4 A m a x = 1 2 3 1 P 2 Also read
Derivative examples
Differentiation of functions
