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Let \( g(x) = f^{-1}(x) \), and \(f'(x) = \dfrac1{1+x^3} \). Find \(g'(x) \).

Note: \(p'(x) \) denotes the derivative of \(p(x) \).

Note by D K 2 years, 2 months ago

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\(\displaystyle f(g(x))=x\) , \(\displaystyle f'(g(x))g'(x)=1\implies \frac{g'(x)}{1+g^3(x)}=1 \implies g'(x)=1+g^3(x)\)

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Note that \[g'(f(x))=1+x^3\\\implies g'(x)=1+(g(x))^3\] To solve this de, use variable separition and partial fractions.

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewest\(\displaystyle f(g(x))=x\) , \(\displaystyle f'(g(x))g'(x)=1\implies \frac{g'(x)}{1+g^3(x)}=1 \implies g'(x)=1+g^3(x)\)

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Note that \[g'(f(x))=1+x^3\\\implies g'(x)=1+(g(x))^3\] To solve this de, use variable separition and partial fractions.

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