Chain Rule 1 (M)

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Chris Knapp

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2 months, 1 week ago

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Chris Knapp 2 months, 1 week ago

Published this.

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Chris Knapp 2 months, 1 week ago

Gave some feedback: Ready to use

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Chris Knapp 1 year, 6 months ago

Created this as a copy of Chain Rule of a polynomial.

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Question statement

Differentiate the following with respect to $x$ :

$y=\simplify{({a}x^{b}+{c})^{{d}}}$

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			Name	Type	Generated Value

a	integer	-8
b	integer	8
c	integer	6
d	integer	5

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random(-8..8 except [-1,0,1])

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Generated value: `integer`

-8

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$\dfrac{dy}{dx} =$

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$\simplify{{a}{b}{d}({a}x^{{b}}+{c})^{{d}-1}x^{{b}-1}}$

Marks: Distractor Message:
$\simplify{-{a}{b}{d}({a}x^{{b}}+{c})^{{d}-1}x^{{b}+1}}$

Marks: Distractor Message:
$\simplify{{a}{b}({a}x^{{b}}+{c})^{{d}}x^{{b}-1}}$

Marks: Distractor Message:
$\simplify{{d}({a}x^{{b}}+{c})^{{d}-1}x^{{b}}}$

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$\simplify{{b}{d}({a}x^{{b}}+{c})^{{d}-1}x^{{b}}}$

Marks: Distractor Message:
$\simplify{{a}{d}({a}{b}x^{{b}-1}+{c})^{{d}-1}}$

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Advice

Let $u(x)=x^{\var{d}}$ and $v(x)=\simplify{{a}x^{{b}}+{c}}$ , then $y=u(v(x))$ .

Using the chain rule, we get

$\dfrac{dy}{dx}= u'(v(x))\times v'(x) = \simplify{{d}({a}x^{b}+{c})^{{d}-1}}\times \simplify{{a}{b}x^{{b}-1}} = \simplify{{d}x^{{b}-1}({a}x^{b}+{c})^{{d}-1}{a}{b}}$

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This question is used in the following exams:

Practice Test by Chris Knapp in Chris's workspace.
Practice Test by Chris Knapp in Chris's workspace.
week24_test1 by Michael Warby in Michael's workspace.
Practice by Chris Knapp in Chris's workspace.

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Chris Knapp 2 months, 1 week ago

Chris Knapp 2 months, 1 week ago

Chris Knapp 1 year, 6 months ago

Error in variable testing condition

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Chain Rule 1 (M)

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Chris Knapp 2 months, 1 week ago

Chris Knapp 2 months, 1 week ago

Chris Knapp 1 year, 6 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`