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$f\left(x\right)\=x^3-3x^2+x+3$$\stackrel{\text{assign as function}}{\to }$$f$

Algebraic Manipulation

Remarks

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

original derivative

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

applied Sum and Difference rules

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

applied Constant rule

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

applied Constant Multiple rule

$3x^2-3\(2x\)\+1$

applied Identity and Power rules

$3x^2-6x\+1$

arithmetic ⇒ final answer

Table 2.3.1(a)   Differentiation of $f\left(x\right)\=x^3-3x^2+x+3$ 

$\frac{\ⅆ}{\ⅆ x}\left(x^3-3 x^2\+x\+3\right)$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Differentiation Steps}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3-3x^2+x+3\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(f_1\left(x\right)+f_2\left(x\right)+f_3\left(x\right)+f_4\left(x\right)\right)=\frac{\ⅆ}{\ⅆx}{f}_1\left(x\right)+\frac{\ⅆ}{\ⅆx}{f}_2\left(x\right)+\frac{\ⅆ}{\ⅆx}{f}_3\left(x\right)+\frac{\ⅆ}{\ⅆx}{f}_4\left(x\right)\\ & & f_1\left(x\right)=x^3\\ & & f_2\left(x\right)=-3x^2\\ & & f_3\left(x\right)=x\\ & & f_4\left(x\right)=3\\ & & \text{This gives:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3\right)+\frac{\ⅆ}{\ⅆx}\left(-3x^2\right)+\frac{\ⅆ}{\ⅆx}x+\frac{\ⅆ}{\ⅆx}3\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{constant}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}3\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}C=0\\ & \text{◦}& \text{This means}\\ & & \frac{\ⅆ}{\ⅆx}3=0\\ & & \text{We can now rewrite the derivative as:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3\right)+\frac{\ⅆ}{\ⅆx}\left(-3x^2\right)+\frac{\ⅆ}{\ⅆx}x\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(-3x^2\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(Cf\left(x\right)\right)=C\frac{\ⅆ}{\ⅆx}f\left(x\right)\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}\left(-3x^2\right)=\left(\mathrm{-3}\right)\cdot \frac{\ⅆ}{\ⅆx}\left(x^2\right)\\ & & \text{We can rewrite the derivative as:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3\right)-3\frac{\ⅆ}{\ⅆx}\left(x^2\right)+\frac{\ⅆ}{\ⅆx}x\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(x^2\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^n\right)=n{x}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=2\cdot x^1\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=2\cdot x\\ & & \text{We can rewrite the derivative as:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3\right)-6x+\frac{\ⅆ}{\ⅆx}x\\ \text{▫}& & \text{5. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}x\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^n\right)=n{x}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}{x}^1=1\cdot x^0\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆ}{\ⅆx}x=1\\ & & \text{We can rewrite the derivative as:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3\right)-6x+1\\ \text{▫}& & \text{6. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(x^3\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^n\right)=n{x}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3\right)=3\cdot x^2\\ & & \text{We can rewrite the derivative as:}\\ & & 3x^2-6x+1\end{array}$

Table 2.3.1(b)   Annotated stepwise solution generated in Maple

                                            or

$\frac{\ⅆ}{\ⅆ x}\left(f\left(x\right)\right)$

Figure 2.3.1(a)   Maple-provided differentiation rule

Figure 2.3.1(b)   User-selected differentiation rule