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May 25th, 2014, 04:47 PM   #1
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Derivation and simplification

I am asked to find the derivative and simplify…

y=(x^3 - 3x^2 + 2√x - 5)/(√x)

Of course at first glance I think to use the product rule but would it be easier to jump straight to logarithmic differentiation?

When I try to use logarithmic differentiation, and I ln both sides etc., I get stuck at dy/dx (or y prime) = (1/√x)((3x^2 - 6x + x^-1/2)(√x) - (1/2)(x^3 - 3x^2 + 2√x - 5)(x^-1/2)/(√x)

I think I am getting lost in the simplification. If someone could help I would greatly appreciate it!
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May 25th, 2014, 06:57 PM   #2
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Hello, OrangeT!

Quote:
Differentiate and simplify: $\displaystyle \:y\:=\:\frac{x^3 - 3x^2 + 2\sqrt{x} - 5}{\sqrt{x}}$

We have: $\displaystyle \:y \;=\;\frac{x^3 - 3x^2 + 2x^{\frac{1}{2}} - 5}{x^{\frac{1}{2}}} \;=\;\frac{x^3}{x^{\frac{1}{2}}} - \frac{3x^2}{x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}}}{x^{\frac{1}{2}}} - \frac{5}{x^{\frac{1}{2}}} $

$\displaystyle \;\;\;$Hence: $\displaystyle \:y \;=\;x^{\frac{5}{2}} - 3x^{\frac{3}{2}} + 2 - 5x^{-\frac{1}{2}}$

Then: $\displaystyle \:y' \;=\;\tfrac{5}{2}x^{\frac{3}{2}} - \tfrac{9}{2}x^{\frac{1}{2}} + \tfrac{5}{2}x^{-\frac{3}{2}} $

$\displaystyle \qquad\quad y' \;=\;\tfrac{1}{2}x^{-\frac{3}{2}}(5x^3 - 9x^2 + 5)$

$\displaystyle \qquad\quad y' \;=\;\frac{5x^3 - 9x^2 + 5x}{2x^{\frac{3}{2}}}$
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May 28th, 2014, 11:45 PM   #3
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Thank you. I understand perfectly. Sometimes I don't see the needed alternative method of simplification. I simply see a fraction and think quotient rule. I will practice this!

Note: I meant to say "quotient" above instead of "product".

Thanks again!
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