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  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, OrangeT! Quote:
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 
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