The formula of product rule is a formula that is used to determine the derivatives of products of two or more functions. Suppose that u(x) and v(x) be differentiable functions. In this way, the product of the functions u(x)v(x) is also differentiable and stated as (uv)′ = u′v + uv′.
The product rule formula is an explanation to the conceptual theory that if the 1st function is multiplied by the derivative of the 2nd in addition to the 2nd function multiplied by the derivative of the 1st function, then the product rule is given. Here we take ‘u’ constant in the 1st term and ‘v’ constant in the 2nd term.
The Product Rule:
\[[f \cdot g]' = f' \cdot g + f \cdot g'\]
Derivation of Product Rule Formula
We prove the above-stated formula of product rule by using the definition of the derivative. For this, we first need to find the increment of the functions uv supposing that the argument alters by Δx:
Δ(uv) = u(x + Δx)v(x + Δx) − u(x)v(x)
u(x + Δx) = u(x) + Δu,v(x + Δx) = v(x) + Δv,
in which, Δu and Δv represent the increments, respectively, of the given functions u and v.
Omitting for brevity the argument x of the respective functions u and v, we can mathematically express the increment Δ(uv) in the below given form:
Δ(uv) = (u + Δu)(v + Δv) − uv = uv + uΔv + vΔu + ΔuΔv − uv = uΔv + vΔu + ΔuΔv.
Now, We head forward to compute the derivative of the product taking into account the properties of limits
(uv)′ = lim Δx→0Δ/Δx = limΔx→0u/Δx = limΔx→0u/Δx + limΔx→0 v/X + limΔx→0 v/Δx + limΔx→0 Δ/Δx⋅limΔx→0Δv.
In the first limit, the function u is not dependent on the increment Δx. Thus, it can also be taken outside the limit sign. A similar theory applies to the function v in the 2nd term.
We are required to separately calculate the limit
limΔx→0Δv=limΔx→0(Δv/Δx⋅Δx) = limΔx→0Δv/Δx⋅limΔx→0Δx = v′⋅0 = 0.
Therefore, the derivative of the product is provided by:
(uv)′ = limΔx→0uΔv/Δx + limΔx→0vΔu/Δx+limΔx→0Δu/Δx⋅limΔx→0Δv=ulimΔx→0Δv/Δx+vlimΔx→0Δu/Δx+limΔx→0ΔuΔx⋅limΔx→0Δv=uv′+vu′+u′⋅0 = u′v + uv′.
Note: The derivative of the product is NOT equivalent to the product of the derivatives!
By this product rule formula, it is also easy to obtain an expression for the derivative of the function kf(x), in which k is a constant:
such that a constant factor can be omitted from the sign of derivative.
Solved Example Using Product Rule Formula
How Will You Differentiate the Function: (x2 + 3) (5x + 4)?
Given function: (a2 + 3)(5a + 4)
u = (x2 + 3) and
v = (5x + 4)
Applying the formula for product rule,
d((x2 +3)(5x+4)/dx = (x2 + 3) d(5x+4)dx + (5x + 4) d(x2+3)dx
= (x2 + 3) 5 + (5x + 4) 2x
= 5x2 + 15 + 10x2 + 8x
= 15x2 + 8x + 15