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Thread: Find derivative.

  1. #1
    Lil
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    Question Find derivative.

    $\displaystyle f(x)=(\sqrt{x}-3x^{3})(2\sqrt{x}+5)$.

    How I do...
    $\displaystyle f'(x)=((\sqrt{x}-3x^{3})(2\sqrt{x}+5))'=$ $\displaystyle (\sqrt{x}-3x^{3})'\cdot (2\sqrt{x}+5)+(\sqrt{x}-3x^{3})\cdot(2\sqrt{x}+5)'=$$\displaystyle (\frac{1}{2\sqrt{x}}-9x^{2})(2\sqrt{x}+5)+(\sqrt{x}-3x^{3})(\frac{2}{2\sqrt{x}}+0)=$ $\displaystyle \frac{2\sqrt{x}}{2\sqrt{x}}+\frac{5}{2\sqrt{x}}-18x^{2}\sqrt{x}-45x^{2}+\frac{\sqrt{x}}{\sqrt{x}}-\frac{3x^{3}}{\sqrt{x}}=$
    $\displaystyle \frac{4\sqrt{x}+5-36x-90x^{2}\sqrt{x}-6x^3}{2\sqrt{x}}$

    I don't know it's good or not. This answer looks strange. Maybe I do mistake? Hmm
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  2. #2
    MHF Contributor

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    Almost. $\displaystyle 18x^2\sqrt{x}\frac{2\sqrt{x}}{2\sqrt{x}}= \frac{32x^3}{2\sqrt{x}}$, not $\displaystyle \frac{36x}{2
    sqrt{x}}$
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