smc-739-60-Chain Rule

Calculus, MET1 2021 VCAA 1a

Differentiate `y = 2e^(-3x)` with respect to `x`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`-6e^(-3x)`

Show Worked Solution

`y`	`=2e^(-3x)`
`dy/dx`	`=-3 xx 2e^(-3x)`
	`=-6e^(-3x)`

Calculus, MET1 2013 VCAA 1b

Let `f(x) = e^(x^2)`.

Find `f^{\prime} (3)`. (3 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`6e^9`

Show Worked Solution

`text(Using Chain Rule:)`

`f^{\prime} (x)`	`= 2xe^(x^2)`
`f^{\prime} (3)`	`= 2 (3) e^((3)^2)`
	`= 6e^9`

Calculus, MET1 2010 VCAA 1b

For `f(x) = log_e (x^2 + 1)`, find `f^{\prime}(2)`. (2 marks)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`4/5`

Show Worked Solution

`text(Using Chain Rule:)`

`f ^{\prime}(x)`	`= (2x)/(x^2 + 1)`
`:. f ^{\prime}(2)`	`= (2(2))/(2^2 + 1)`
	`= 4/5`

Calculus, MET1 2009 ADV 2b

Let `y=ln(3x^3 + 2)`.

Find `dy/dx`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`2e^x(e^x + 1)`

Show Worked Solution

	`y`	`=ln(3x^3 + 2)`
	`(dy)/(dx)`	`=(3*3x^2)/(3x^3 + 2)`
		`=(9x^2)/(3x^3 + 2)`

Calculus, MET1 2020 VCAA 1b

Evaluate `f^{\prime}(1)`, where `f: R -> R, \ f(x) = e^(x^2-x + 3)`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`e^3`

Show Worked Solution

`f(x)`	`= e^(x^2-x + 3)`
`f^{\prime}(x)`	`= (2x-1)e^(x^2-x + 3)`
`f^{\prime}(1)`	`= (2-1)e^(1-1 + 3)`
	`= e^3`

Calculus, MET1 2015 ADV 11e

Differentiate `(e^x + x)^5`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`5(e^x + 1)(e^x + x)^4`

Show Worked Solution

`y`	`= (e^5 + x)^5`
`(dy)/(dx)`	`= 5(e^x + x)^4 xx d/(dx)(e^x + x)`
	`= 5(e^x + x)^4 xx (e^x + 1)`
	`= 5(e^x + 1)(e^x + x)^4`

Calculus, MET1 2009 ADV 2a

Differentiate `(e^x + 1)^2` with respect to `x`. (2 marks)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`2e^x(e^x + 1)`

Show Worked Solution

	`y`	`= (e^x + 1)^2`
	`(dy)/(dx)`	`= 2(e^x + 1)^1 xx d/(dx) (e^x + 1)`
		`= 2e^x(e^x + 1)`

Calculus, MET1 2016 VCAA 1b

Let `f(x) = x^2e^(5x)`.

Evaluate `f^{\prime}(1)`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`7e^5`

Show Worked Solution

`text(Using Product Rule:)`

`(fg)^{\prime}`	`= f^{\prime}g + fg^{\prime}`
`f^{′}(x)`	`= 2xe^(5x) + 5x^2 e^(5x)`
`f^{′}(1)`	`= 2(1)e^(5(1)) + 5(1)^2 e^(5(1))`
	`= 7e^5`

Calculus, MET2 2009 VCAA 7 MC

For `y = e^(2x) cos (3x)` the rate of change of `y` with respect to `x` when `x = 0` is

`0`
`2`
`3`
`– 6`
`– 1`

Show Answers Only

`B`

Show Worked Solution

`y`	`= e^(2x) cos (3x)`
`dy/dx`	`=e^(2x) xx -3sin(3x) + 2e^(2x) xx cos (3x)`
	`=e^(2x)(-3sin(3x) + 2cos(3x))`

`text(When)\ \ x = 0,`

`dy/dx= 2`

`=> B`

Calculus, MET2 2011 VCAA 4 MC

The derivative of `log_e(2f(x))` with respect to `x` is

`(f′(x))/(f(x))`
`2(f′(x))/(f(x))`
`(f′(x))/(2f(x))`
`log_e(2f′(x))`
`2log_e(2f′(x))`

Show Answers Only

`A`

Show Worked Solution

`text(Chain Rule:)`

`text(If)\ \ h(x)`	`= f(g(x))`
`h′(x)`	`= f′(g(x)) xx g′(x)`

`d/(dx)(log_e(2f(x)))`	`= 1/(2f(x)) xx 2f′(x)`
	`= (f′(x))/(f(x))`

`=> A`