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 notrole="math" localid="1648599845313" $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=\left(\frac{df}{dx}\right)\left(\frac{dg}{dx}\right)$.

Short 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 localid="1648643313626" $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=\left(\frac{df}{dx}\right)\left(\frac{dg}{dx}\right)$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-by-step solution

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 ddx(x-3)

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

The derivative of a function is defined as :-

localid="1648642284940" $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

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 ddx(2x+1)

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

The derivative of a function is defined as :-

role="math" localid="1648642563904" $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

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 ddx(x-3)(2x+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 :-

role="math" localid="1648643040763" $$\begin{gathered} \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{\left[\left(x+h-3\right)\left(2\left(x+h\right)+1\right)\right]-\left[\left(x-3\right)\left(2x+1\right)\right]}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{\left[\left(x+h-3\right)\left(2x+2h+1\right)\right]-\left[(x-3)\left(2x+1\right)\right]}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{2x^2+2xh+x+2xh+2h^2+h-6x-6h-3-2x^2-x+6x+3}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{2xh+2xh+2h^2+h-6h}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim }\frac{h(2x+2x+2h+1-6)}{h} \\ \Rightarrow \frac{d}{dx}(x-3)(2x+1)=\underset{h\to 0}{\lim (}2x-6+2x+1+2h) \\ \Rightarrow \frac{d}{dx}(x-3\left)\right(2x+1)=2x-6+2x+1 \\ \Rightarrow \frac{d}{dx}(x-3\left)\right(2x+1)=2(x-3)+(2x+1) \end{gathered}$$

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 :-

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