Higher Order derivatives

Higher order Differentiation is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of differentiation have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

In this article, we will cover the concept of the Differentiation of a Function with another Function and the higher-order derivative of a Function. This concept falls under the broader category of Calculus, a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last five years of the JEE Main exam (from 2013 to 2023), a total of nine questions have been asked on this concept, including one in 2020, two in 2021, two in 2022, and four in 2023.

Higher Order Derivative of a Function

The derivative of a function is itself a function, so we can find the derivative of a derivative. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, a fourth derivative, and so on.

Collectively, these are referred to as higher-order derivatives. The notation for the higher-order derivatives of $y=f(x)$ can be expressed in any of the following forms:

$
\begin{aligned}
& f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x), f^{(4)}(x), \ldots, f^{(n)}(x) \\
& y^{\prime}, y^{\prime \prime}(x), y^{\prime \prime \prime}(x), y^{(4)}(x), \ldots, y^{(n)}(x) \\
& \frac{d y}{d x}, \frac{d^2 y}{d x^2}, \frac{d^3 y}{d x^3}, \frac{d^4 y}{d x^4}, \ldots, \frac{d^n y}{d x^n}
\end{aligned}
$

Note:

It is interesting to note that the notation for $\frac{d^2 y}{d x^2}$ may be viewed as an attempt to express $\frac{d}{d x}\left(\frac{d y}{d x}\right)_{\text {more compactly. }}$
Also $\frac{d}{d x}\left(\frac{d}{d x}\left(\frac{d y}{d x}\right)\right)=\frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=\frac{d^3 y}{d x^3}$
$\frac{d^2 y}{d x^2} \neq\left(\frac{d y}{d x}\right)^2$
$\frac{d^2 y}{d x^2} \neq\left(\frac{\frac{d^2 y}{d t^2}}{\frac{d^2 x}{d t^2}}\right)$
$\frac{d^2 y}{d x^2} \neq \frac{1}{\frac{d^2 x}{d y^2}}$

Solved Examples

Example 1: If $y^2+\log _e\left(\cos ^2 x\right)=y, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then:
[JEE Main 2020]
1) $y^{\prime \prime}(0)=0$
2) $\left|y^{\prime}(0)\right|+\left|y^{\prime \prime}(0)\right|=1$
3) $\left|y^{\prime \prime}(0)\right|=2$
4) $\left|y^{\prime}(0)\right|+\left|y^{\prime \prime}(0)\right|=3$

Solution:

$
\begin{aligned}
& y^2+\log _e\left(\cos ^2 x\right)=y \\
& y^2+2 \log _e(\cos x)=y \\
& 2 y y^{\prime}-2 \frac{\sin x}{\cos x}=y^{\prime} \\
& 2 y y^{\prime}-2 \tan x=y^{\prime} \\
& \therefore y^{\prime}(0)=0 \\
& 2\left(y^{\prime}\right)^2+2 y y^{\prime \prime}-2 \sec ^2 x=y^{\prime \prime} \\
& \therefore\left|y^{\prime \prime}(0)\right|=2
\end{aligned}
$

Hence, the answer is the option 3.

Example 2: Let $f: S \rightarrow S_{\text {where }} S=(0, \infty)$ be twice differentiable function such that $f(x+1)=x f(x)$. If $g: S \rightarrow R_{\text {be defined as }} g(x)=\log _e f(x)$ then the value of $\left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right|_{\text {is equal to: [JEE Main 2021] }}$
1) $\frac{205}{144}$
2) $\frac{197}{144}$
3) $\frac{187}{144}$
4) 0

Solution:

$
\begin{aligned}
& \ln f(x+1)=\ln (x f(x)) \\
& \ln f(x+1)=\ln x+\ln f(x) \\
& \Rightarrow \quad g(x+1)=\ln x+g(x) \\
& \Rightarrow \quad g(x+1)-g(x)=\ln x \\
& \Rightarrow \quad g^{\prime \prime}(x+1)-g^{\prime \prime}(x)=-\frac{1}{x^2}
\end{aligned}
$
Put $x=1,2,3,4$
$
\begin{aligned}
& g^{\prime \prime}(2)-g^{\prime \prime}(1)=-\frac{1}{1^2} \quad \ldots(1) \\
& g^{\prime \prime}(3)-g^{\prime \prime}(2)=-\frac{1}{2^2} \quad \ldots(2) \\
& g^{\prime \prime}(4)-g^{\prime \prime}(3)=-\frac{1}{3^2} \quad \ldots(3) \\
& g^{\prime \prime}(5)-g^{\prime \prime}(4)=-\frac{1}{4^2} \quad \ldots(4)
\end{aligned}
$
Add all the equations we get

$
\begin{aligned}
& g^{\prime \prime}(5)-g^{\prime \prime}(1)=-\frac{1}{1^2}-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2} \\
& \left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right|=\frac{205}{144}
\end{aligned}
$

Hence, the answer is the option 1.

Example 3: If

$
\cos ^{-1}\left(\frac{y}{2}\right)=\log _{\mathrm{e}}\left(\frac{x}{5}\right)^5,|y|<2
$

then?
[JEE Main 2022]
1) $x^2 y^{\prime \prime}+x y^{\prime}-25 y=0$
2) $x^2 y^{\prime \prime}-x y^{\prime}-25 y=0$
3) $x^2 y^{\prime \prime}-x y^{\prime}+25 y=0$
4) $x^2 y^{\prime \prime}+x y^{\prime}+25 y=0$

Solution:

$
\begin{aligned}
& \cos ^{-1} \frac{\mathrm{y}}{2}=5 \log _{\mathrm{e}} \frac{\mathrm{x}}{5} \\
& \Rightarrow \frac{\mathrm{y}}{2}=\cos \left(5 \log _{\mathrm{e}} \frac{\mathrm{x}}{5}\right) \\
& \Rightarrow \mathrm{y}^{\prime}=-2 \sin \left(5 \log _{\frac{\mathrm{x}}{5}}^5\right) \times \frac{5}{\mathrm{x} / 5} \times \frac{1}{5} \\
& =\frac{-10 \sin \left(5 \log _{\mathrm{e}} \frac{\mathrm{x}}{5}\right)}{\mathrm{x}}
\end{aligned}
$

$\begin{aligned} & \Rightarrow y^{\prime \prime}=\frac{-x \times 10 \cos \left(5 \log _e \frac{x}{5}\right) \times \frac{5}{x / 5} \times \frac{1}{5}+10 \sin \left(5 \log _e \frac{x}{5}\right)}{x^2} \\ & \Rightarrow x^2 y^{\prime \prime}=-50 \cos \left(5 \log _e \frac{x}{5}\right)+10 \sin \left(5 \log _e \frac{x}{5}\right) \\ & \Rightarrow x^2 y^{\prime \prime}=-25 y-x y^{\prime} \\ & \Rightarrow x^2 y^{\prime \prime}+x y^{\prime}+25 y=0\end{aligned}$ṁ

Hence the answer is the option (4).

Example 4: Let

$
f^1(x)=\frac{3 x+2}{2 x+3}, x \in \mathbf{R}-\left\{\frac{-3}{2}\right\}
$
For $n \geq 2$, define $f^n(x)=f^1$ of $f^{n-1}(x)_{\text {If }} f^5(x)=\frac{a x+b}{b x+a}, \operatorname{gcd}(a, b)=1$, then $\mathrm{a}+$ $b$ is equal to:
[JEE Main 2023]
1) 3125
2) 3130
3) 3135
4) 3140

Solution:

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{3 \mathrm{x}+2}{2 \mathrm{x}+3}, \mathrm{x} \in \mathrm{R}-\left\{-\frac{3}{2}\right\} \\ & \mathrm{f}^2(\mathrm{x})=\mathrm{f}^{\prime} 0 \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}^{\prime}\left(\frac{3 \mathrm{x}+2}{2 \mathrm{x}+3}\right) \\ & =\frac{3\left(\frac{3 \mathrm{x}+2)}{2 \mathrm{x}+3)}\right)+2}{2\left(\frac{3 \mathrm{x}+2}{2 \mathrm{x}+3}\right)+3} \\ & =\frac{9 x+6+4 x+6}{6 x+4+6 x+9} \\ & f^2(x)=\frac{13 x+12}{12 x+13} \\ & f^3(x)=f^{\prime} o f^2(x) \\ & =f^{\prime}\left(\frac{13 x+12}{12 x+13}\right)\end{aligned}$

$\begin{aligned} & =\frac{3\left(\frac{13 x+12}{12 x+13}\right)+2}{2\left(\frac{13 x+12}{12 x+13}\right)+3} \\ & =\frac{39 x+36+24 x+26}{26 x+24+36 x+39} \\ & f^3(x)=\frac{63 x+62}{62 x+63} \\ & f^4(x)=f^1\left(\frac{63 x+62}{62 x+63}\right) \\ & =\frac{3\left(\frac{63 x+62}{62 x+63}\right)+2}{2\left(\frac{63 x}{62 x}+\frac{62}{63}\right)+2} \\ & f^4(x)=\frac{313 x+312}{312 x+313} \\ & f^5(x)=f^1\left(\frac{313 x+312}{312 x+313}\right)\end{aligned}$

$\begin{aligned} & =\frac{3\left(\frac{313 x+312}{312 x+313}\right)+2}{2\left(\frac{313 x+312}{312 x+313}\right)+3} \\ & \mathrm{f}^5(\mathrm{x})=\frac{1563 \mathrm{x}+1562}{1562 \mathrm{x}+1563} \\ & \because \mathrm{a}=1563, \mathrm{~b}=1562 \\ & \mathrm{a}+\mathrm{b}=3125\end{aligned}$

Hence, the answer is the option(1).

Example 5: If $f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in \mathbb{R}$ then:
[JEE Main 2023]
1) $f(1)+f(2)+f(3)=f(0)$
2) $2 \mathrm{f}(0)-\mathrm{f}(1)+\mathrm{f}(3)=\mathrm{f}(2)$
3) $3 f(1)+f(2)=f(3)$
4) $f(3)-f(2)=f(1)$

Solution:

$
\begin{aligned}
& f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3) \\
& f(x)=x^3-a x^2+b x-c \\
& f^{\prime}(x)=3 x^2-2 a x+b \\
& f^{\prime \prime}(x)=6 x-2 a
\end{aligned}
$

$\begin{aligned} & f^{\prime \prime \prime}(x)=6 \\ & f^{\prime \prime \prime}(3)=6 \\ & f^{\prime}(1)=3-2 a+b=a \Rightarrow 3 a=b+3 \\ & f^{\prime \prime}(2)=12-2 a=b \Rightarrow 2 a=12-b \text { Page } \\ & \quad a=3, b=6 \\ & f^{\prime \prime \prime}(3)=6=c \\ & f(x)=x^3-3 x^2+6 x-6 \\ & f(0)=-6 \quad f(2)=2 \\ & f(1)=-2 \quad f(3)=12\end{aligned}$

Hence, the answer is the option 2.

Summary: Differentiation of a Function with another Function and the higher-order derivative of a Function is an important concept in Calculus. It provides a deep understanding of how the functions interact and change. It is very helpful in practical applications for physics, economics, etc. The chain rule is used to find the derivative. The derivative can be found a lot of times like the second, third, and fourth derivative.