Question

How do I find f'(x) for f(x)=7(x^(4/5))-8(x^(6/7))?

I got as far as 5.6(x^(-1/5))-6.857(x^(-1/7)) but this does not appear to be either the final form of the answer or the correct answer according to the system I do homework on. I'm confused about the power rules, and would appreciate any help!

Answer 1

`f^{'}(x)\ \ =\ \ \frac{28}{5x^{\frac{1}{5}}}-\frac{48}{7x^{\frac{1}{7}}}`

Explanation:

`f(x) = 7x^{\frac{4}{5}}-8x^{\frac{6}{7}`

`f^'(x) = \frac{d}{dx}(7x^{\frac{4}{5}}-8x^{\frac{6}{7}})`

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`\text{Apply the Sum/Difference of derivative Rule}:`

`\quad (f(x)\pm g(x))^'=f^'(x)\pm g^'(x)`

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`=\frac{d}{dx}(7x^{\frac{4}{5}})-\frac{d}{dx}(8x^{\frac{6}{7}})`

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Take the constant out `\ \ \ \` `(a\cdot f(x))^'=a\cdot f^'(x)` `\ \ \ \` and apply the power rule `\ \ \ \` `\quad \frac{d}{dx}(x^a)=a\cdot x^{a-1}` `\ \ \ \` to get:

`=7\frac{d}{dx}(x^{\frac{4}{5}})-8\frac{d}{dx}(x^{\frac{6}{7}})`

`=7\cdot \frac{4}{5}x^{\frac{4}{5}-1}-8\cdot \frac{6}{7}x^{\frac{6}{7}-1}`

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Simplify the exponents and then rewrite the terms by using the exponent rule `\ \ \` `a^{-b}=\frac{1}{a^b}`

`=\frac{28}{5}x^{-\frac{1}{5}}\ \ -\ \ \frac{48}{7}x^{-\frac{1}{7}}`

`=\frac{28}{5x^{\frac{1}{5}}}-\frac{48}{7x^{\frac{1}{7}}}`

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That's it!