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 May 25th, 2014, 04:47 PM #1 Member   Joined: Apr 2014 From: Japan Posts: 37 Thanks: 3 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!  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: $\displaystyle \:y' \;=\;\tfrac{5}{2}x^{\frac{3}{2}} - \tfrac{9}{2}x^{\frac{1}{2}} + \tfrac{5}{2}x^{-\frac{3}{2}}$

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

$\displaystyle \qquad\quad y' \;=\;\frac{5x^3 - 9x^2 + 5x}{2x^{\frac{3}{2}}}$ May 28th, 2014, 11:45 PM #3 Member   Joined: Apr 2014 From: Japan Posts: 37 Thanks: 3 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! Tags calculus 1, calculus1, derivation, simplification Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post greg1313 Algebra 1 July 8th, 2012 09:51 PM Solarmew Applied Math 8 April 27th, 2012 01:08 PM tupi157 Calculus 1 October 22nd, 2011 02:37 AM bluz Calculus 7 June 30th, 2011 03:15 AM eric3353 Advanced Statistics 0 August 19th, 2008 03:46 PM

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