Q 5 - Calculus

Question

Use the definition of the derivative (or exercises done

previously in this section) to find (a) $\frac{d}{dx}(x-3)$, (b) $\frac{d}{dx}(2x+1)$, and (c) $\frac{d}{dx}\left(\right(x-3\left)\right(2x+1\left)\right)$. Use your answers to make a conjecture as to whether or not  $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=\left(\frac{df}{dx}\right)\left(\frac{dg}{dx}\right)$.

Step-by-Step Solution

Answer

(a) $\frac{d}{dx}(x-3)=1$

(b) $\frac{d}{dx}(2x+1)=2$

(c) $\frac{d}{dx}(x-3)(2x+1)=2(x-3)+(2x+1)$

From the above results we can conclude that the given conjecture $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=\left(\frac{df}{dx}\right)\left(\frac{dg}{dx}\right)$ is not true. That is :-

$\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)\ne \left(\frac{df}{dx}\right)\left(\frac{dg}{dx}\right)$

Step 1. Given Information

We have given the following three functions :-

$\left(a\right) \left(x-3\right), \left(b\right) (2x+1), \left(c\right) \left(\left(x-3\right)\left(2x+1\right)\right)$

We have to find derivative of these functions.

By using these results we have to find that the following conjecture is true or not :-

$\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=\left(\frac{df}{dx}\right)\left(\frac{dg}{dx}\right)$

Step 2. Part (a) To find the value of d d x ( x - 3 )

We have to find the derivative of function $f\left(x\right)=x-3$.

The derivative of a function  is defined as :-

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

Put all the values, then we have :-

$$\begin{gathered} \frac{d}{dx}(x-3)=\underset{h\to 0}{\lim }\frac{x+h-3-x+3}{h} \\ \Rightarrow \frac{d}{dx}(x-3)=\underset{h\to 0}{\lim }\frac{h}{h} \\ \Rightarrow \frac{d}{dx}(x-3)=\underset{h\to 0}{\lim }1 \\ \Rightarrow \frac{d}{dx}(x-3)=1 \end{gathered}$$

Step 2. Part (b) To find the value of d d x ( 2 x + 1 )

We have to find the derivative of function $f\left(x\right)=2x+1$.

The derivative of a function  is defined as :-

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

Put all the values, then we have :-

$$\begin{gathered} \frac{d}{dx}(2x+1)=\underset{h\to 0}{\lim }\frac{2(x+h)+1-(2x+1)}{h} \\ \Rightarrow \frac{d}{dx}(2x+1)=\underset{h\to 0}{\lim }\frac{2x+2h+1-2x-1}{h} \\ \Rightarrow \frac{d}{dx}(2x+1)=\underset{h\to 0}{\lim }\frac{2h}{h} \\ \Rightarrow \frac{d}{dx}(2x+1)=\underset{h\to 0}{\lim }2 \\ \Rightarrow \frac{d}{dx}(2x+1)=2 \end{gathered}$$

Step 4. Part (c) To find the value of d d x ( x - 3 ) ( 2 x + 1 )

We have to find the derivative of the function $f\left(x\right)=(x-3)(2x+1)$.

The derivative of a function  is defined as :-

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

Put all the values, then we have :-

$$\begin{gathered} \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{\left[\left(x+h-3\right)\left(2\left(x+h\right)+1\right)\right]-\left[\left(x-3\right)\left(2x+1\right)\right]}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{\left[\left(x+h-3\right)\left(2x+2h+1\right)\right]-\left[(x-3)\left(2x+1\right)\right]}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{2x^2+2xh+x+2xh+2h^2+h-6x-6h-3-2x^2-x+6x+3}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{2xh+2xh+2h^2+h-6h}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{h(2x+2x+2h+1-6)}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim (}2x-6+2x+1+2h) \\ \Rightarrow \frac{d}{dx}(x-3\left)\right(2x+1)=2x-6+2x+1 \\ \Rightarrow \frac{d}{dx}(x-3\left)\right(2x+1)=2(x-3)+(2x+1) \end{gathered}$$

Step 5. Conclusion for conjecture

In previous steps, we find that :-

$$\begin{gathered} \frac{d}{dx}(x-3)=1 ..........\left(1\right) \\ \frac{d}{dx}(2x+1)=2 ..........\left(2\right) \\ \frac{d}{dx}(x-3)(2x+1)=2(x-3)+(2x+1) ......\left(3\right) \end{gathered}$$

From the above equations we can see that :-

$\frac{d}{dx}(x-3)(2x+1)\ne \frac{d}{dx}(x-3)\times \frac{d}{dx}(2x+1)$

Then for the following conjecture :-

$\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=\left(\frac{df}{dx}\right)\left(\frac{dg}{dx}\right)$ is not true. That is :-

$\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)\ne \left(\frac{df}{dx}\right)\left(\frac{dg}{dx}\right)$.