Answer:
(i)
[tex]\displaystyle \large{f'(x) = 6x - 2}[/tex]
(ii/(a))
[tex] \displaystyle \large{y' = 10 {x}+ 3 {x}^{2} - 28{x}^{3} } [/tex]
(ii/(b))
[tex] \displaystyle \large{y' = 20 {x}^{ 3} + 90{x}^{2} + 120x +40}[/tex]
Step-by-step explanation:
( 1 )
For first question, we have to use first principal or limit definition to find the derivative of 3x^2-2x.
Limit Definition of Derivative
[tex] \displaystyle \large{f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} }[/tex]
Our f(x) is 3x^2-2x.
Our f(x+h) is 3(x+h)^2-2(x-h)
[tex] \displaystyle \large{f'(x) = \lim_{h \to 0} \frac{[3 {(x + h)}^{2} - 2(x + h)]- (3 {x}^{2} - 2x) }{h} }[/tex]
Expand the expressions and simplify:-
[tex] \displaystyle \large{f'(x) = \lim_{h \to 0} \frac{[3 ( {x}^{2} + 2xh + {h}^{2}) - 2x - 2h]- 3 {x}^{2} + 2x}{h} } \\ \displaystyle \large{f'(x) = \lim_{h \to 0} \frac{3 {x}^{2} + 6xh + 3{h}^{2} - 2x - 2h - 3 {x}^{2} + 2x}{h} } \\ \displaystyle \large{f'(x) = \lim_{h \to 0} \frac{6xh + 3 {h} ^{2} - 2h}{h} }[/tex]
Cancel all h's.
[tex] \displaystyle \large{f'(x) = \lim_{h \to 0} \frac{6xh + 3 {h} ^{2} - 2h}{h} } \\ \displaystyle \large{f'(x) = \lim_{h \to 0} 6x + 3h - 2}[/tex]
Substitute h = 0
[tex] \displaystyle \large{f'(x) = \lim_{h \to 0} 6x + 3h - 2} \\ \displaystyle \large{f'(x) = \lim_{h \to 0} 6x + 3(0) - 2} \\ \displaystyle \large{f'(x) = \lim_{h \to 0} 6x- 2} \\ \displaystyle \large{f'(x) = 6x - 2}[/tex]
Therefore, the derivative of 3x^2-2x is 6x-2 by first principal.
( 2 )
For the second question, there are two sub-questions. Assume that you can use any formulas to differentiate instead of using first principal.
(a)
We are given the function:
[tex] \displaystyle \large{y = 5 {x}^{2} + {x}^{3} - 7 {x}^{4} }[/tex]
Our formula for derivative of polynomial function is:-
[tex] \displaystyle \large{y = a {x}^{n} \longrightarrow y' = na {x}^{n - 1} } [/tex]
Apply the formula:-
[tex] \displaystyle \large{y = 5 {x}^{2} + {x}^{3} - 7 {x}^{4} } \\ \displaystyle \large{y' = 2(5) {x}^{2 - 1} + 3 {x}^{3 - 1} - 4(7) {x}^{4 - 1} } \\ \displaystyle \large{y' = 10 {x}+ 3 {x}^{2} - 28{x}^{3} } [/tex]
If you want it in factored form then:-
[tex]\displaystyle \large{y' = 10 {x}+ 3 {x}^{2} - 28{x}^{3} } \\ \displaystyle \large{y' = x(10 + 3 {x}- 28{x}^{2} )} [/tex]
Alternative
You can factor the expression then apply product rules.
Product Rules
[tex] \displaystyle \large{y = f(x)g(x) \longrightarrow y' = f' (x)g(x) + g' (x)f(x)}[/tex]
Therefore:-
[tex]\displaystyle \large{y = 5 {x}^{2} + {x}^{3} - 7 {x}^{4} } \\ \displaystyle \large{y = x(5 {x} + {x}^{2} - 7 {x}^{3} )} \\ \displaystyle \large{y' = x'(5x + x^{2} - {7x}^{3} ) + x(5x + {x}^{2} - 7 {x}^{3} )'} \\ \displaystyle \large{y' = 5x + x^{2} - {7x}^{3} + x(5+ 2 {x} - 21 {x}^{2}) } \\ \displaystyle \large{y' = 5x + x^{2} - {7x}^{3} + 5x + 2 {x}^{2} - 21 {x}^{3} } \\ \displaystyle \large{y' = 3 {x}^{2} + 10x - 28 {x}^{3} }[/tex]
( b )
For our second sub-question, you can apply product rules instead of expanding in and differentiate.
[tex] \displaystyle \large{y = 5 x {(x + 2)}^{3} } \\ \displaystyle \large{y' = 5 x ' {(x + 2)}^{3} + 5x((x + 2) ^{3})' } \\ \displaystyle \large{y' = 5 {(x + 2)}^{3} + 5x(3(x + 2) ^{2}) }[/tex]
Recall cubic formula:
[tex] \displaystyle \large{ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} }[/tex]
Therefore:-
[tex] \displaystyle \large{y' = 5 ( {x}^{3} + 6 {x}^{2} + 12x + 8) + 5x(3( {x}^{2} + 4x + 4) } \\ \displaystyle \large{y' = 5 {x}^{3} + 30 {x}^{2} + 60x + 40+ 5x(3{x}^{2} + 12x + 12) } \\ \displaystyle \large{y' = 5 {x}^{3} + 30 {x}^{2} + 60x + 40 + 15 {x}^{3} + {60x}^{2} + 60x } \\ \displaystyle \large{y' = 20 {x}^{ 3} + 90{x}^{2} + 120x +40}[/tex]
