Differentiation (Harder powers - negative and fractions) 4a

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Ben McGovern

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Christian Lawson-Perfect 2 years, 6 months ago

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Ben McGovern 4 years, 8 months ago

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Name	Status	Author	Last Modified
Differentiation (Harder powers - negative and fractions) 4a	draft	Ben McGovern	27/09/2022 09:01
Differentiation (Harder powers - negative and fractions) 4b	draft	Ben McGovern	27/09/2022 09:01

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Question statement

If we have a function of the form $y=f(g(x))$ , sometimes described as a function of a function, to calculate its derivative we need to use the chain rule:

$\frac{dy}{dx} = \frac{du}{dx} \times \frac{dy}{du}.$

This can be split up into stages:

Let $u=g(x)$ ;
Rewrite $y$ in terms of $u$ , such that $y=f(u)$ ;
Calculate $\frac{du}{dx}$ and $\frac{dy}{du}$ ;
Write $\frac{dy}{dx}$ as a product of $\frac{du}{dx}$ and $\frac{dy}{du}$ ;
Make sure $\frac{dy}{dx}$ is only in terms of $x$ . Ensure any $u$ terms have been replaced using the initial substitution.

Calculate the derivative of $y=\simplify[all,fractionNumbers]{{k}({a}*x^{m}+{b})^{n}}$ .

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			Name	Type	Generated Value

a	integer	2
b	integer	9
n	integer	-5
k	rational	-4/7
m	integer	3

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Following the process of calculating the derivative of a function using the chain rule, we must first identify $g(x)$ . Since the function is of the form $y=f(g(x))$ , we are looking for the 'inner' function:

So, for $y=\simplify[all,fractionNumbers]{{k}({a}*x^{m}+{b})^{n}}$ ,

$g(x)=\simplify[all, fractionNumbers, unitFactor]{{a}x^{m}+{b}}.$

If we now set $u=g(x)$ , we can rewrite $y$ in terms of $u$ such that $y=f(u)$ :

$y=\simplify[all, fractionNumbers,unitFactor]{{k}u^{n}}.$

Next, we calculate the two derivatives $\frac{du}{dx}$ and $\frac{dy}{du}$ :

$\frac{du}{dx}=\simplify[all,fractionNumbers]{{a*m}x^{m-1}}, \quad \frac{dy}{du}=\simplify[all, fractionNumbers, unitFactor]{{k*n}*u^{n-1}}.$

Plugging these into the chain rule:

$\begin{split} \frac{dy}{dx} &= \frac{du}{dx} \times \frac{dy}{du}, \\&=\simplify[all,fractionNumbers]{{a*m}x^{m-1}} \times\simplify[all, fractionNumbers, unitFactor]{{k*n}u^{n-1}}, \\ &=\simplify[all, fractionNumbers, unitFactor]{{k*n*a*m}*x^{m-1}u^{n-1}}. \end{split}$

Finally, we need to express $\frac{dy}{dx}$ only in terms of $x$ , so we must replace the $u$ terms using the initial substitution $u=\simplify[all, fractionNumbers, unitFactor]{{a}x^{m}+{b}}$ :

$\frac{dy}{dx} = \simplify[all, fractionNumbers, unitFactor]{{k*n*a*m}*x^{m-1}({a}x^{m}+{b})^{n-1}}.$

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Differentiation (Harder powers - negative and fractions) 4a

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Christian Lawson-Perfect 2 years, 6 months ago

Ben McGovern 4 years, 8 months ago

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Differentiation (Harder powers - negative and fractions) 4a

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Christian Lawson-Perfect 2 years, 6 months ago

Ben McGovern 4 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

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Generated value: `integer`