Q. 56 - Calculus

Question

Use the definition of the derivative to find the derivatives described in Exercises 55-58.

Find ${\left.\frac{d}{dx}\left(2x^3\right)\right|}_3,{\left.\frac{d^2}{d{x}^2}\left(2x^3\right)\right|}_1,{\left.\frac{d^3}{d{x}^3}\left(2x^3\right)\right|}_{-2}$

Step-by-Step Solution

Answer

The values are $54,12,12$ respectively

Step 1. Given information

Given function $y=2x^3$

Step 2: Use the definition of derivative and calculate

Calculating, we get

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{2(x+h{)}^3-2x^3}{h} \\ \underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}==\underset{h\to 0}{\lim }\frac{2\left(x^3+h^3+3x^{2}h+3x{h}^2\right)-2x^3}{h} \\ =\underset{h\to 0}{\lim }\frac{2x^3+2h^3+6x^{2}h+6x{h}^2-2x^3}{h} \\ =\underset{h\to 0}{\lim }\frac{2h^3+6x^{2}h+6x{h}^2}{h} \\ =6x^2 \\ 6{\left(3\right)}^2=6\times 9 \\ =54 \end{gathered}$$

Step 3: Calculate other derivatives

Calculating, we get

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{g(x+h)-g\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{6(x+h{)}^2-6x^2}{h} \\ \underset{h\to 0}{\lim }\frac{g(x+h)-g\left(x\right)}{h}==\underset{h\to 0}{\lim }\frac{6\left(x^2+h^2+2xh\right)-6x^2}{h} \\ =\underset{h\to 0}{\lim }\frac{6h^2+12xh}{h} \\ =12x \\ 12\left(1\right)=12 \end{gathered}$$

Step 4: Calculating other derivatives

Calculating, we get

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{m(x+h)-m\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{12(x+h)-12x}{h} \\ \underset{h\to 0}{\lim }\frac{m(x+h)-m\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{12x+12h-12x}{h} \\ =\underset{h\to 0}{\lim }\frac{12h}{h} \\ =12 \end{gathered}$$