Leibnitz Theorem formula proof and solved examples
In Mathematics, the Leibnitz theorem or Leibniz integral rule for derivation comes under the integral sign. It is named after the famous scientist Gottfried Leibniz. Thus, the theorem is basically designed for the derivative of the antiderivative. Basically, the Leibnitz theorem is used to generalise the product rule of differentiation. It states that if there are two functions let them be a(x) and b(x) and if they both are differentiable individually, then their product a(x). b(x) is also n times differentiable.
This theorem basically refers to the process through which one can find the derivative of an antiderivative. It is also known as successive differentiation. According to the proposition, the derivative on the nth order of the product of two functions can be expressed with the help of a formula. The formula for the above-mentioned theorem is as follows:
\[ (uv)^{n} = \sum_{i=0}^{n} \left(\begin{array}{c}n\\ i\end{array}\right) u^{n-i} v^{i} \]
In the above expression, \[ \left(\begin{array}{c}n\\ i\end{array}\right) \] represents the total number of i-combinations.
Hence, the Leibnitz theorem/formula for the nth derivative has been mentioned above.
Derivation of the Leibnitz Theorem/Formula
If A and B are the functions of x1, then dn(AB)/dxn = [nC0.An.B] + [nC1.An-1.B1] + [nC2.An-2.B2] + [nCr.An-r.Br] +...........+ [nCn.A.Bn]. Thus, the theorem will be proved by induction.
Step 1: By actual differentiation, we already know that
(AB)1 = A1.B + A.B1 (AB)2
= (A2.B + A1.B1) + (A1.B1 + A.B2)
= A2.B + 2A1.B1 + A.B2
= A2.B + 2A1.B1+ A.B2
= A2.B + 2C1 A1.B1 + A.B2
Hence, the theorem holds true for n = 1,2.
Step 2: We assume that the theorem is true for a particular value of n say k, so we have:
(AB)k = [Ak.B] + [kC1 Ak−1.B1] + [kC2 Ak−2.B2] + ...+ [kCr1Ak−r+1.Br−1] + [kCr Ak−r.Br + kCk A.Bk]
Differentiating both sides, we get
(AB)k+1 = [Ak+1.B + Ak.B1] + [kC1 Ak.B1 + kC1 . Ak−1.B2] +...+ [kC2 Ak−1.B2 + kC2 Ak−2.B3] + ...+ [kCr−1 Ak−r+2.Br−1] + [kCr−1 Ak−r+1.Br] + [kCr Ak−r+1.Br + kCr Ak−r.Br+1] +...[kCk A1.Bk + kCk A.Bk+1 (AB)k+1]
= [Ak+1.B +(1 + kC1) Ak.B1] + (kC1 + kC2) Ak−1.B2 +...+ ( kCr−1+ kCr ) Ak−r+1.Br +...+ kCk A.Bk+1 (kCr−1 + kCr = (k+1)Cr) (AB)k+1.
= Ak+1.B + (k+1)C1 Ak.B1 + (k+1)C2Ak−1.B2 +...+ (k+1)Cr Ak−r+1.Br +...+ (k+1)Ck+1 A.Bk+1
Thus, the theorem is true for n = k+1, i.e., it also holds true for the next higher integral value of k. Given that, we have observed that the theorem is true for n = 2, therefore, the theorem is true for (n = 2 + 1), i.e., n = 3, and, therefore, further true for n = 4 and so on. Hence, the theorem is true for all positive real values of n.
Solved Examples
Let's do some sample question solving:
Q1: If y = x3 eax, find yn , using Leibnitz theorem.
Let \[ u = e^{ax}\] , \[v = x^{3}\] . Now, \[ u_{n} = a^{n}e^{ax} \]
By Leibnitz’s Theorem ,
\[ y_{n} = a^{n} e^{ax} x^{3} + \left(\begin{array}{c}n\\ 1\end{array}\right) a^{n-1}e^{ax} 3x^{2} + \left(\begin{array}{c}n\\ 2\end{array}\right) a^{n-2}e^{ax} 6x + \left(\begin{array}{c}n\\ 3\end{array}\right) a^{n-3}e^{ax} 6 \]
= \[ e^{ax} a^{n-2} (a^{3}x^{3} + 3na^{2}x^{2} + 3n (n-1) ax + n(n-1) (n-2) ) \]
Consider a function in two variables x and y, i.e.,
z=f(x,y)
Let us consider the integral of z with respect to x, from a to b, i.e.,
\[I = \int_{a}^{b} f (x,y)dx \]
For this integration, the variable is only x and not y. y is essentially a constant for the integration process. Therefore, after we have evaluated the definite integral and put in the integration limits, y will still remain in the expression of I. This means that I is a function of y.
\[I (y) = \int_{a}^{b} f (x,y) dx \]… (1)
This relation (1) can be differentiated with respect to y as follows:
\[ I’ (y) = \frac{d}{dy} \left ( \int_{a}^{b} f(x,y)dx\right ) \]
=\[ \int_{a}^{b} \frac{\partial f(x,y)}{\partial{y}} dx \]
Where
\[ \frac{\partial f(x,y)}{\partial{y}} \]
stands for the partial derivative of f(x,y) with respect to y, that is, the derivative of f(x,y) w.r.t. y, treating x as a constant.
So as per the rules if you want to find a derivative in the nth order of the product of two functions, then this formula of Leibnitz is going to help you. Sometimes we refer to it as the differential under the integral sign.
FAQs on Leibnitz Theorem for Nth Derivative of a Product
1. What is Leibnitz theorem in calculus?
The Leibnitz theorem gives the formula for the nth derivative of the product of two functions. If y = u(x)v(x), then the nth derivative is:
dⁿ(uv)/dxⁿ = Σ [ⁿCᵏ · dⁿ⁻ᵏu/dxⁿ⁻ᵏ · dᵏv/dxᵏ], where k = 0 to n.
- ⁿCᵏ is the binomial coefficient.
- u and v are differentiable functions.
- It is an extension of the product rule to higher-order derivatives.
2. What is the formula for Leibnitz rule for nth derivative?
The formula for the Leibnitz rule for the nth derivative of a product uv is:
dⁿ(uv)/dxⁿ = ⁿC₀u⁽ⁿ⁾v + ⁿC₁u⁽ⁿ⁻¹⁾v′ + ⁿC₂u⁽ⁿ⁻²⁾v″ + ... + ⁿCₙuv⁽ⁿ⁾.
- u⁽ⁿ⁾ denotes the nth derivative of u.
- v⁽ⁿ⁾ denotes the nth derivative of v.
- The coefficients follow the binomial expansion pattern.
3. How do you prove Leibnitz theorem?
The proof of Leibnitz theorem is done using mathematical induction.
- Step 1: Verify for n = 1 (ordinary product rule).
- Step 2: Assume the formula is true for n = m.
- Step 3: Differentiate the assumed result once more.
- Step 4: Use binomial coefficient properties to obtain the formula for n = m + 1.
4. Can you give an example of Leibnitz theorem?
Yes, Leibnitz theorem can be used to find higher derivatives of products like y = x²eˣ.
Using the formula:
dⁿ(x²eˣ)/dxⁿ = eˣ[x² + 2nx + n(n−1)].
- u = x², so higher derivatives become zero after second order.
- v = eˣ, whose derivatives remain eˣ.
- Substitute into the Leibnitz formula and simplify.
5. What is the difference between product rule and Leibnitz theorem?
The product rule finds the first derivative of a product, while Leibnitz theorem finds the nth derivative.
- Product rule: (uv)' = u'v + uv'
- Leibnitz theorem: General formula for dⁿ(uv)/dxⁿ
- Leibnitz rule extends the product rule to higher orders.
6. When is Leibnitz theorem used in differentiation?
Leibnitz theorem is used when finding the higher-order derivatives of a product of two functions.
- Common in Taylor and Maclaurin series expansions.
- Useful in solving differential equations.
- Applied when repeated differentiation of products is required.
7. What are the conditions for applying Leibnitz rule?
The main condition for applying Leibnitz rule is that both functions must be sufficiently differentiable up to order n.
- u(x) and v(x) must have derivatives up to nth order.
- The derivatives must exist and be continuous in the interval considered.
- n must be a positive integer.
8. How do binomial coefficients appear in Leibnitz theorem?
The binomial coefficients (ⁿCᵏ) appear because the formula follows the binomial expansion pattern.
- Each term contains ⁿCᵏ.
- The derivatives are distributed between u and v.
- The structure resembles (a + b)ⁿ expansion.
9. How do you find the second derivative using Leibnitz theorem?
Using Leibnitz theorem for n = 2, the second derivative of uv is:
d²(uv)/dx² = u''v + 2u'v' + uv''.
- This matches the formula with coefficients 1, 2, 1.
- It follows the binomial pattern (1, 2, 1).
- It can also be derived by applying the product rule twice.
10. What are common mistakes when using Leibnitz theorem?
Common mistakes in applying Leibnitz theorem include incorrect coefficients and missing terms.
- Forgetting the binomial coefficient ⁿCᵏ.
- Stopping derivatives too early.
- Ignoring that some higher derivatives may become zero.
- Incorrect algebraic simplification after substitution.
