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.
If u = f(x,y).g(x,y), then
ux = ∂x∂u = g(x,y)∂x∂f+f(x,y)∂x∂g
And, uy = ∂y∂u = g(x,y)∂y∂f+f(x,y)∂y∂g
If u = g(x,y)f(x,y), where g(x,y) = 0 then,
ux = [g(x,y)]2g(x,y)∂x∂f−f(x,y)∂x∂g
And, uy = [g(x,y)]2g(x,y)∂y∂f−f(x,y)∂y∂g
If u = [f(x,y)]2 then, partial derivative of u with respect to x and y defined as
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:
us = ∂x∂u.∂s∂x+∂y∂u.∂s∂y
and ut = ∂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
Given function is f(x, y) = tan(xy) + sin x
Derivative of a function with respect to x is given as follows:
fx = ∂x∂f = ∂x∂[tan(xy)+sinx]
= sec2 (xy). y + cos x
= y sec2(xy) + cos x
Now, Derivative of a function with respect to y. So, x is constant
fy = ∂y∂f = ∂y∂[tan(xy)+sinx]
= 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.
We know, dtdf=∂x∂fdtdx+∂y∂fdtdy
Then, ∂x∂f = 2, ∂y∂f = 3, dtdx = 1, dtdy = 2t
So, we get
dtdf = 2.1 + 3.(2t) = 2 + 3t
Question 3: If f=x2(y–z)+y2(z–x)+z2(x–y), prove that ∂x∂f + ∂y∂f + ∂z∂f+0
To find ∂x∂f ‘y and z’ are held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’