nth Derivative: Definition, Examples, Leibnitz Theorem

The nth derivative of a function f(x) is obtained by by successively differentiating f(x) n times with respect to x. The Leibnitz theorem is a powerful tool to find it for a product function. In this post, we will learn how to find n-th derivative of a function with examples.

Definition of nth Derivative

The first order derivative of f(x) is denoted by $f'(x)$, and it is given by the limit: $f'(x)=\lim\limits_{h \to 0}$ $\dfrac{f(x+h)-f(x)}{h}$.

The n-th order derivative of f(x) is denoted by f⁽ⁿ⁾(x) and it is obtained by differentiating f^(n-1)(x) with respect to x. So we have:

$f^{(n)}(x)=\lim\limits_{h \to 0}$ $\dfrac{f^{(n-1)}(x+h)-f^{(n-1)}(x)}{h}$.

Notation: The nth derivative of $y$ is usually denoted by $y_n$.

Examples of nth Derivative

Q1: Find the nth order derivative of f(x)=xn.

Answer: Here f(x)=xⁿ.

By the power rule of differentiation, we have that

$f'(x)=nx^{n-1}$ $f^{”}(x)=n(n-1)x^{n-2}$ $f^{”’}(x)=n(n-1)(n-2)x^{n-3}$

So the kth order derivative of f(x)=xⁿ is given by

$f^{(k)}(x)=n(n-1)…(n-k+1)x^{n-k}$.

Thus, to obtain the nth order derivative, let us put k=n here. As a result we obtain that

$f^{(n)}(x)=n(n-1)…(n-n+1)x^{n-n}$

⇒ $f^{(n)}(x)=n(n-1)…1$ as we know x⁰=1.

⇒ $f^{(n)}(x)=n!$

So the nth order derivative of xⁿ is equal to n!.

Remark: The mth order derivative of f(x)=xⁿ is given by

Q2: Find the nth order derivative of f(x)=sinx.

Answer: Here f(x)=sinx.

The first derivative is given by

$f'(x)=\cos x$

⇒ $f'(x)=\sin(\frac{\pi}{2}+x)$

In a similar way, we obtain that

$f^{”}(x)=\cos(\frac{\pi}{2}+x)$ $f^{”}(x)= \sin(\frac{\pi}{2}+\frac{\pi}{2}+x)$ using the formula sin(π/2 +θ) = cosθ. ⇒ $f^{”}(x) = \sin(2 \cdot \frac{\pi}{2}+x)$

Also, we have:

$f^{(3)}(x)=\cos(2 \cdot \frac{\pi}{2}+x)$ $f^{(3)}(x)= \sin(\frac{\pi}{2}+2\cdot \frac{\pi}{2}+\frac{\pi}{2}+x)$ as we know sin(π/2 +θ) = cosθ. ⇒ $f^{(3)}(x) = \sin(3 \frac{\pi}{2}+x)$

Observing the patterns, we see that $f^{(n)}(x) = \sin(n \cdot \frac{\pi}{2}+x)$.

So the nth derivative of sinx is equal to sin(nπ/2 +x).

Note:

1. The nth derivative of sin(ax+b) is equal to aⁿ sin(nπ/2 +ax+b).
2. In a similar way, the derivative of cosx is equal to cos(nπ/2 +x). Also, the nth derivative of cos(ax+b) is equal to aⁿ cos(nπ/2 +ax+b).

Q3: Find the nth order derivative of f(x)=$\dfrac{1}{x}$.

Answer: Here f(x)= 1/x = x^-1.

So f'(x) = -1 ⋅ x-1-1 = -1 ⋅ x-2 f”(x) = -1 ⋅ -2 x-2-1 = -1 ⋅ -2 x-3 f”'(x) = -1 ⋅ -2 ⋅ -3 x-3-1 = -1 ⋅ -2 ⋅ -3 x-4

Therefore, we observe that

f⁽ⁿ⁾(x) = -1 ⋅ -2 ⋅ -3 ⋅⋅⋅ -n x^-n-1 = (-1)ⁿ 1⋅ 2 ⋅ 3⋅n x^-(n+1)

⇒ f⁽ⁿ⁾(x) = $\dfrac{(-1)^n n!}{x^{n+1}}$.

So the nth derivative of 1/x is equal to (-1)ⁿ n! x^-(n+1).

Q4: Find the nth order derivative of f(x)=log x.

Answer: Here f(x)=log x.

Differentiating with respect to x, we get that

f'(x) = $\dfrac{1}{x}$.

Differentiating it (n-1) times w.r.t x using the rule in Q3, we obtain the nth order derivative of f(x) = 1/x which is given below:

f⁽ⁿ⁾(x) = $\dfrac{(-1)^{n-1} (n-1)!}{x^n}$.

So the nth order derivative of logx is equal to (-1)^n-1 (n-1)! x^-n.

Leibnitz Theorem

Statement: For two continuously differentiable functions u and v, the nth order derivative of the product uv is given by the following formula:

(uv)_n = u_nv + ⁿc₁ u_n-1v₁ + ⁿc₂ u_n-2v₂ + … + uv_n.

Application

Q5: Find the nth derivative of y = ex logx.

Answer:

y = e^x logx

Put u=e^x and v=log x.

∴ u_n = e^x and v_n = $\dfrac{(-1)^{n-1}(n-1)!}{x^n}$.

Therefor, by Leibnitz’s theorem, the nth order derivative of y = e^x logx is given by

y_n = e^x ⋅ log x + ⁿc₁ ⋅ e^x ⋅$\dfrac{1}{x}$ + ⁿc₂ ⋅ e^x $\left( -\dfrac{1}{x^2}\right)$ + … + e^x $\dfrac{(-1)^{n-1}(n-1)!}{x^n}$

⇒ y_n = e^x { log x + nx^-1 + ⁿc₂ x^-2 + … + (-1)^n-1 (n-1)! x^-n }.

Q6: If y = tan-1x, then show that (1+x2)yn+1 +2nxyn +n(n-1)yn-1 = 0.

Answer:

Differentiating the both sides with respect to x, we have that

y₁ = $\dfrac{1}{1+x^2}$

⇒ (1+x²)y₁ = 1

Let us now differentiate n-times with respect to x. This gives us that

[y₁(1+x²)]_n = (1)_n

⇒ y_n+1(1+x²) + ⁿc₁ y_n ⋅2x + ⁿc₂ y_n-1 ⋅2 = 0

⇒ (1+x²)y_n+1 + 2nxy_n + n(n-1)y_n-1 = 0, this completes the proof.

Q7: If y = em sin-1x, then prove that (1-x2)yn+2 – (2n+1)xyn+1 -(n2+m2)yn = 0, where yn denotes the nth derivative of y.

Answer:

y = e^{m sin-1x}.

Differentiate both sides with respect to x. So we have that

y₁ = $e^{m \sin^{-1}x} \times \dfrac{m}{\sqrt{1-x^2}}$

Cross-multiplying, we get that

y₁²(1-x²) = m²y².

If we differentiate again with respect to x, we obtain that

2y₁ y₂(1-x²) +y₁² ⋅(-2x) = m²⋅2yy₁

Cancelling 2y₁, we have: y₂ (1-x²) – xy₁ = m²y.

Now, we differentiate using Leibnitz’s rule, it follows that

yn+2 (1-x2) + nc1 yn+1 ⋅(-2x) + nc1 yn ⋅(-2) – {yn+1x + nc1 yn ⋅1} = m2yn ⇒ (1-x2)yn+2 – 2nxyn+1 – n(n-1)yn – xn+1 – nyn = m2yn

Therefore, (1-x²)y_n+2 – (2n+1)xy_n+1 -(n²+m²)y_n = 0.