smc-745-50-Chain Rule

Calculus, MET2 2020 VCAA 7 MC

If `f(x)=e^(g(x^(2)))`, where `g` is a differentiable function, then `f^(')(x)` is equal to

`2xe^(g(x^(2)))`
`2xg(x^(2))e^(g(x^(2)))`
`2xg^(')(x^(2))e^(g(x^(2))`
`2xg^(')(2x)e^(g(x^(2)))`
`2xg^(')(x^(2))e^(g(2x))`

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`C`

Show Worked Solution

`f(x)=e^(g(x^2))`

`text{Using the chain rule (twice):}`

`f^{‘}(x)`	`=d/dx[g(x^2)] * e^(g(x^2))`
	`=2xg^{‘}(x^2)e^(g(x^2))`

`=> C`

Calculus, MET1 2021 VCAA 1a

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

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`-6e^(-3x)`

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`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)

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`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)

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`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 2020 VCAA 1b

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

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`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 2008 VCAA 1b

Let `f(x) = xe^(3x)`. Evaluate `f^{prime}(0)`. (3 marks)

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`1`

Show Worked Solution

`text(Using Product Rule:)`

`(gh)^{prime} = g^{prime}h + gh^{prime}`

`f^{prime}(x)`	`= x(3e^(3x)) + 1 xx e^(3x)`
`:.f^{prime}(0)`	`= 0 + e^0`
	`= 1`

Calculus, MET1 2015 ADV 11e

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

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`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, 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))`

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`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`

Calculus, MET1 2007 VCAA 2b

Let `g(x) = log_e(tan(x))`. Evaluate `g^{prime}(pi/4)`. (2 marks)

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`g^{prime}(pi/4) = 2`

Show Worked Solution

`g(x) = log_e (tan(x))`

`text(Using Chain Rule:)`

`g^{prime}(x) = (sec^2(x))/(tan(x))`

`text(When)\ \ x = pi/4,`

`g^{prime}(pi/4)`	`= (sec^2(pi/4))/(tan (pi/4))`
	`=1/(1/sqrt2)^2`
	`=1/(1/2)`
	`=2`