Derivative of a Function wrt Another Function

Function

$A$ and $B$ are two non-empty sets, then a relation from $A$ to $B$ is said to be a function if each element $x$ in $A$ is assigned a unique element $f(x)$ in $B$, and it is written as
$f: A \rightarrow B$ and read as $f$ is a mapping from $A$ to $B$.

Derivative of a Function

Let $f$ be defined on an open interval $I \subseteq$ containing the point $x_0$, and suppose that $\lim _{\Delta x \rightarrow 0} \frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}$ exists. Then $f$ is said to be differentiable at $x_0$ and the derivative of $f$ at $x_0$, denoted by $f^{\prime}\left(x_0\right)$, is given by

$
f^{\prime}\left(x_0\right)=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}
$

For all $x$ for which this limit exists,
$f^{\prime}(x)=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$ is a function of $x$.

In addition to $f^{\prime}(x)$, other notations are used to denote the derivative of $y=f(x)$. The most common notations are $f^{\prime}(x), \frac{d y}{d x}, y^{\prime}, \frac{d}{d x}[f(x)], D_x[y]$ or $D y$ or $y_1$. Here $\frac{d}{d x}$ or $D$ is the differential operator.

Properties of derivative of a function

1. The derivative of sum of two functions is equal to the sum of their derivatives.

(i.e)., $\frac{d}{dx}[f(x)+g(x)] = \frac{d}{dx}[f(x)]+\frac{d}{dx}[g(x)]$

2. The derivative of differnce betweeen two functions is equal to the difference between their derivatives.

(i.e)., $\frac{d}{dx}[f(x)+g(x)] = \frac{d}{dx}[f(x)]+\frac{d}{dx}[g(x)]$

3. The derivative of the product of two functions is given by

$\frac{d}{dx}[f(x)g(x)] = (\frac{d}{dx}f(x))g(x)+f(x)(\frac{d}{dx}g(x))$

4. The derivative of the quotient of two functions is given by

$\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{(\frac{d}{dx}f(x))g(x)+f(x)(\frac{d}{dx}g(x))}{(g(x))^2}$

Derivative of a function with respect to another function

Derivative of a function with respect to another function is differentiating two functions with a dependent vaiable. toChain rule can be usedo find the derivative of a function with respect to another.

Chain rule

Let $f(x)$ and $g(x)$ be two functions. To differentiate $f(x)$ with respect to $g(x)$, Let $u=f(x)$ and $v=g(x)$.

$\left(\frac{\mathrm{d} u}{\mathrm{~d} v}\right)=\frac{\frac{\mathrm{d} u}{\mathrm{~d} x}}{\frac{\mathrm{d} x}{\mathrm{~d} v}}$ where $\frac{\mathrm{d} x}{\mathrm{~d} v}\neq0$

Recommended Video Based on Derivative of a Function with respect to Another Function

Solved Examples Based on Derivative of a Function with respect to Another Function

Example 1:Find the second order derivative if x and y are given by $x=a \sin t$ and $y=a \operatorname{cost}$
1) $
\frac{1}{a} \cos ^3 t
$

2) $
-\frac{1}{a} \cos ^3 t
$

3) $
\frac{1}{a} \sec ^3 t
$

4) $
-\frac{1}{a} \sec ^3 t
$

Solution:

Differentiating the function implicitly with respect to " $x$ ";

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{d y}{d t}}{\frac{\mathrm{~d} x}{d t}}=\frac{-a \sin t}{a \cos t}=-\frac{\sin t}{\cos t}
$

Again differentiating with respect to " $x$ ";

$
\begin{aligned}
& \frac{d^2 y}{d x^2}=\frac{\mathrm{d}}{\mathrm{~d} x}\left(\frac{d y}{d x}\right)=\frac{\mathrm{d}}{\mathrm{~d} t}\left(-\frac{\sin t}{\cos t}\right) \frac{d t}{d x} \\
& \frac{d^2 y}{d x^2}=-\left(\sec ^2 t\right) \times \frac{1}{a \cos t}=-\frac{\sec ^2 t}{a \cos t}=-\frac{1}{a} \sec ^3 t
\end{aligned}
$

Hence, the answer is the option 4.

Example 2: Differentiate $\cos \left(a x^2+b x+c\right)$ with respect to $sin$ $\left(l x^2+m x+n\right)$
1) $
\frac{-(2 a x+b) \cos \left(a x^2+b x+c\right)}{(2 l x+m) \sin \left(l x^2+m x+n\right)}
$

2) $
\frac{-(2 a x+b) \sin \left(a x^2+b x+c\right)}{(2 l x+m) \cos \left(l x^2+m x+n\right)}
$

3) $
\frac{-(2 l x+m) \cos \left(l x^2+m x+n\right)}{(2 a x+b) \sin \left(a x^2+b x+c\right)}
$

4) $
\frac{-(2 l x+m) \sin \left(l x^2+m x+n\right)}{(2 a x+b) \cos \left(a x^2+b x+c\right)}
$

Solution:

Let $f(x)=\cos \left(a x^2+b x+c\right)$ and $g(x)=\sin \left(l x^2+m x+n\right)$

$
\begin{aligned}
& \frac{d f}{d g}=\frac{f^{\prime}(x)}{g^{\prime}(x)} \\
& f^{\prime}(x)=-\sin \left(a x^2+b x+c\right)(2 a x+b) \\
& g^{\prime}(x)=\cos \left(l x^2+m x+n\right)(2 l x+m) \\
& \frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{-(2 a x+b) \sin \left(a x^2+b x+c\right)}{(2 l x+m) \cos \left(l x^2+m x+n\right)}
\end{aligned}
$

Hence, the answer is the option 2.

Example 3: Find the derivative of $\log _{10}(x)$ with respect to $x^3$.
1) $
\frac{1}{2 x^2\left(\log _e 10\right)}
$

2) $
\frac{1}{2 x^3\left(\log _e 10\right)}
$

3) $
\frac{1}{3 x^3\left(\log _e 10\right)}
$

4) $
\frac{1}{3 x^2\left(\log _e 10\right)}
$

Solution:

Let $u =\log _{10}(x)$ and $v=x^3$
Also,$\frac{d u}{d x}=\frac{1}{x \log _e(10)}$ and $\frac{d v}{d x}=3 x^2$
$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$
$= \frac{\frac{1}{x \log _c(10)}}{3 x^2}=\frac{1}{3 x^3 \log _e(10)}$

Hence, the answer is the option 3.

Example 4: If $y=\frac{\sqrt{\left(1+t^2\right)}-\sqrt{\left(1-t^2\right)}}{\sqrt{\left(1+t^2\right)}+\sqrt{\left(1-t^2\right)}}$ and $x=\sqrt{\left(1-t^4\right)}$ then $\frac{d y}{d x}$ is equal to
1) $
\frac{-1}{t^2\left\{1+\sqrt{\left.\left(1-t^4\right)\right\}}\right.}
$

2) $
\frac{\left\{\sqrt{\left(1-t^4\right)}-1\right\}}{t^6}
$

3) $
\frac{1}{t^2\left\{1+\sqrt{\left.\left(1-t^4\right)\right\}}\right.}
$

4) $
\frac{1-\sqrt{\left(1-t^4\right)}}{t^6}
$

Solution:
$
\begin{aligned}
& y=\frac{\left(\sqrt{1+t^2}-\sqrt{1-t^2}\right)^2}{\left(\sqrt{1+t^2}\right)^2-\left(\sqrt{1-t^2}\right)^2} \\
& =\frac{2-2 \sqrt{1-t^4}}{2 t^2}=\frac{1-\sqrt{1-t^4}}{t^2} \\
& \frac{d y}{d t}=\frac{t^2\left\{0-\frac{1}{2 \sqrt{1-t^4}} \times-4 t^3\right\}-\left\{1-\sqrt{1-t^4}\right\} 2 t}{t^4} \\
& =\frac{2 t^5}{\sqrt{1-t^4}}-2 t\left\{1-\sqrt{1-t^4}\right\} \\
& =\frac{2\left\{1-\sqrt{1-t^4}\right\}}{t^3 \sqrt{1-t^4}} \\
& \frac{d x}{d t}=\frac{-2 t^3}{\sqrt{1-t^4}} \\
& \frac{d y}{d x}=\frac{\sqrt{1-t^4}-1}{t^6}=\frac{-1}{t^2\left\{1+\sqrt{1-t^4}\right\}}
\end{aligned}
$

Hence, the answer is the option 1.

Example 5: If $y=x^3+e^x$, then find $\frac{d^2 y}{d x^2}$.
1) $
-\frac{\left(5 x+e^x\right)}{\left(3 x^2+e^x\right)^3}
$

2) $
-\frac{\left(6 x+e^x\right)}{\left(3 x^2+e^x\right)^3}
$

3) $
-\frac{\left(6 x+e^x\right)}{\left(3 x^2+e^x\right)^2}
$

4) $
-\frac{\left(5 x+e^x\right)}{\left(3 x^2+e^x\right)^2}
$

Solution:

Given that, $y=x^3+e^x$

$\begin{aligned} & \frac{d y}{d x}=3 x^2+e^x \\ & \Rightarrow \frac{d x}{d y}=\frac{1}{3 x^2+e^x} \\ & \Rightarrow \frac{d^2 x}{d y^2}=-\frac{1}{\left(3 x^2+e^x\right)^2} \frac{d}{\mathrm{~d} y}\left(3 x^2+e^x\right) \\ & \Rightarrow \frac{d^2 x}{d y^2}=-\frac{1}{\left(3 x^2+e^x\right)^2}\left[6 x+e^x\right] \frac{d x}{d y} \\ & \Rightarrow \frac{d^2 x}{d y^2}=-\frac{6 x+e^x}{\left(3 x^2+e^x\right)^2} \times \frac{1}{\left(3 x^2+e^x\right)} \\ & \Rightarrow \frac{d^2 x}{d y^2}=-\frac{6 x+e^x}{\left(3 x^2+e^x\right)^3}\end{aligned}$

Hence, the answer is the option (2)

Summary

Derivative of a Function with respect to another 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 of a function with respect to another.