Q 4 - Calculus

Question

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

Step-by-Step Solution

Answer

(a) $\frac{d}{dx}\left(3x\right)=3$

(b) $\frac{d}{dx}\left(x^2\right)=2x$

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

From the above results we have that the conjecture :-

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

Step 1. Given Information

We have given the following three functions :- 

$\left(a\right) 3x, \left(b\right) x^2, \left(c\right) 3x+x^2$.

We have to find derivative of these functions.

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

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

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

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

The derivative of a function $f\left(x\right)$ 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 :-

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

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

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

We know that the derivative of a function $f\left(x\right)$ is defied 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 :- 

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

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

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

The derivative of a function $f\left(x\right)$ 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 :- 

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

Step 5. Conclusion for conjecture

In previous steps, we find that :-

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

By comparing $\left(1\right),\left(2\right),\left(3\right)$ equations, we can write :-

$\frac{d}{dx}(3x+x^2)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^2\right)$

If we generalize this conclusion for any functions, then we have the following conjecture :-

$\frac{d}{dx}\left(f\right(x)+g(x\left)\right)=\frac{d}{dx}\left(f\right(x\left)\right)+\frac{d}{dx}\left(g\right(x\left)\right)$.

So the conjecture is true.