This problem aims to familiarize us with **different methods** to solve a **differential.** The concept required to cater to this **problem** mostly relates to **ordinary differential equations.** We define an **ordinary differential equation** or most commonly known as **ODE,** as an equation that has one or **additional functions** of a **single independent variable** given with their derivatives. On the other hand, an **equation** that includes a **function** more than a **single derivative** is known as a **differential equation.** But as we speak of **ODE,** the term **ordinary** is employed for the **derivative** of **one independent variable.**

The **rules** that are going to use in this **problem** are the **product rule, quotient rule,** and **chain rule.**

Whenever a **function** contains **another function** within it, we **differentiate** that function with the help of the **chain rule.** It is given as:

\[ f(g(x)) \]

The **derivative** can then be taken as:

\[ \dfrac{d}{dx}(f(g(x)) = f'(g(x))\cdot g'(x) \]

\[ \dfrac{dy}{dx} = \dfrac{dy}{du}\cdot \dfrac{du}{dx} \]

The **product rule** as it says is the **derivative** of **two functions** that are arithmetically being **multiplied,** given as:

\[ \dfrac{d}{dx}(f \cdot g) = f\cdot \dfrac{dg}{dx} + g\cdot \dfrac{df}{dx} \]

Whereas the **quotient rule** applies to the **functions** that are in the form of a **fraction,** given as:

\[ \dfrac{d}{dx} \{\dfrac{f(x)}{g(x)}\} = \dfrac{g\cdot \dfrac{df}{dx} – f\cdot \dfrac{dg}{dx}}{g^2}\]

## Expert Answer

We are given the following **information:**

\[ f(5) = 1,\space f'(5) = 6\]

\[ g(5) = -3,\space g'(5) = 2\]

First, we are going to **find** $(f(x)\cdot g(x))$ using the **product rule:**

\[ \dfrac{d}{dx}(f\cdot g) = f\dfrac{dg}{dx} + g\dfrac{df}{dx} \]

\[ \dfrac{d}{dx}(f(5)g(5)) = f(5)g'(5) + g(5)f'(5) \]

\[ \dfrac{d}{dx}(f(5)g(5)) = 1\times 2 + (-3)\times 6 \]

\[ \dfrac{d}{dx}(f(5)g(5)) = -16 \]

**Next,** we are going to **find** $(\dfrac{f(x)}{g(x)})’$ using the **quotient rule:**

\[ \dfrac{d}{dx} \{\dfrac{f(5)}{g(5)}\} = \dfrac{g(5)f'(5) – f(5)g'(5)}{g(5)^2} \]

\[ (\dfrac{f(5)}{g(5)})’ = \dfrac{(-3)\times 6 – 1\times 2}{(-3)^2} \]

\[ (\dfrac{f(5)}{g(5)})’ = \dfrac{-18 – 2}{9} \]

\[ (\dfrac{f(5)}{g(5)})’ = \dfrac{-20}{9} \]

And **finally,** we are going to **find** $(\dfrac{g(x)}{f(x)})’$ using the **quotient rule:**

\[ \dfrac{d}{dx} \{\dfrac{g(5)}{f(5)}\} = \dfrac{f(5)g'(5) – g(5)f'(5)}{f(5)^2} \]

\[ (\dfrac{g(5)}{f(5)})’ = \dfrac{1\times 2 – (-3)\times 6}{1^2} \]

\[ (\dfrac{g(5)}{f(5)})’ = \dfrac{2 + 20}{1} \]

\[ (\dfrac{g(5)}{f(5)})’ = 20 \]

## Numerical Result

**Part a:** $\dfrac{d}{dx}(f(5)g(5)) = -16$

**Part b:** $(\dfrac{f(5)}{g(5)})’ = \dfrac{-20}{9}$

**Part c:** $(\dfrac{g(5)}{f(5)})’ = 20$

## Example

Given that $f(3)=1$, $f'(3)=8$, $g(3)=-6$, and $g'(3)=2$. Find the **following differentials,** $(fg)'(3)$, $(f/g)'(3)$ and $(g/f)'(3)$.

According to the **statement,** we are **given:**

\[ f(3) = 1,\space f'(3) = 8\]

\[ g(3) = -6,\space g'(3) = 2\]

**First, finding** $(f(x)\cdot g(x))$:

\[ \dfrac{d}{dx}(f\cdot g) = f\dfrac{dg}{dx} + g\dfrac{df}{dx}\]

\[ \dfrac{d}{dx}(f(3)g(3)) = f(3)g'(3) + g(3)f'(3) \]

\[ (f(3)g(3))’ = 1\times 2 + (-6)\times 8 \]

\[ (f(3)g(3))’ = -46 \]

**Next,** finding $(\dfrac{f(x)}{g(x)})’$:

\[ \dfrac{d}{dx} \{\dfrac{f(3)}{g(3)}\} = \dfrac{g(3)f'(3) – f(3)g'(3)}{g(3)^2} \]

\[ (\dfrac{f(3)}{g(3)})’ = \dfrac{(-6)\times 8 – 1\times 2}{(-6)^2} \]

\[ (\dfrac{f(3)}{g(3)})’ = \dfrac{-48 – 2}{36} \]

\[ (\dfrac{f(3)}{g(3)})’ = \dfrac{-25}{18} \]

**And finally,** $(\dfrac{g(x)}{f(x)})’$:

\[ \dfrac{d}{dx} \{\dfrac{g(3)}{f(3)}\} = \dfrac{f(3)g'(3) – g(3)f'(3)}{f(3)^2} \]

\[ (\dfrac{g(3)}{f(3)})’ = \dfrac{1\times 2 – (-6)\times 8}{1^2} \]

\[ (\dfrac{g(5)}{f(5)})’ = \dfrac{2 + 48}{1} \]

\[ (\dfrac{g(5)}{f(5)})’ = 50 \]