Question

Help me with this one.............​

Answer 1

(i)

`\displaystyle \large{f'(x) = 6x - 2}`

(ii/(a))

`\displaystyle \large{y' = 10 {x}+ 3 {x}^(2) - 28{x}^(3) }`

(ii/(b))

`\displaystyle \large{y' = 20 {x}^( 3) + 90{x}^(2) + 120x +40}`

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

`\displaystyle \large{f'(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h) }`

Our f(x) is 3x^2-2x.

Our f(x+h) is 3(x+h)^2-2(x-h)

`\displaystyle \large{f'(x) = \lim_(h \to 0) \frac{[3 {(x + h)}^(2) - 2(x + h)]- (3 {x}^(2) - 2x) }{h} }`

Expand the expressions and simplify:-

`\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} }`

Cancel all h's.

`\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}`

Substitute h = 0

`\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}`

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:

`\displaystyle \large{y = 5 {x}^(2) + {x}^(3) - 7 {x}^(4) }`

Our formula for derivative of polynomial function is:-

`\displaystyle \large{y = a {x}^(n) \longrightarrow y' = na {x}^(n - 1) }`

Apply the formula:-

`\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) }`

If you want it in factored form then:-

`\displaystyle \large{y' = 10 {x}+ 3 {x}^(2) - 28{x}^(3) } \\ \displaystyle \large{y' = x(10 + 3 {x}- 28{x}^(2) )}`

Alternative

You can factor the expression then apply product rules.

Product Rules

`\displaystyle \large{y = f(x)g(x) \longrightarrow y' = f' (x)g(x) + g' (x)f(x)}`

Therefore:-

`\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) }`

( b )

For our second sub-question, you can apply product rules instead of expanding in and differentiate.

`\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)) }`

Recall cubic formula:

`\displaystyle \large{ {(x + y)}^(3) = {x}^(3) + 3 {x}^(2) y + 3x {y}^(2) + {y}^(3) }`

Therefore:-

`\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}`