paulo1913 10 Posted May 15, 2009 Share Posted May 15, 2009 How do I differentiate this? (x^2-3x+2)(sqrt x) What i did was this: (x^2-3x+2)(1/2x^-1/2)+(2x-3)(sqrt x) But what is the next step? Link to post Share on other sites

feign_ignorence 10 Posted May 15, 2009 Share Posted May 15, 2009 How do I differentiate this? (x^2-3x+2)(sqrt x) What i did was this: (x^2-3x+2)(1/2x^-1/2)+(2x-3)(sqrt x) But what is the next step? Expand and simplify Link to post Share on other sites

paulo1913 10 Posted May 15, 2009 Author Share Posted May 15, 2009 How do you expand using square roots? I haven't come across that before Link to post Share on other sites

feign_ignorence 10 Posted May 16, 2009 Share Posted May 16, 2009 How do you expand using square roots? I haven't come across that before (x^2-3x+2)(1/2x^-1/2)+(2x-3)(sqrt x) [(x^4/2)-(3x^2/2)+(2)][(1/2)(x^-1/2)]+[2x^-6/2][x^1/2] Its a bit hard to show using text, but its just basic exponent multiplication. Same base = add the exponents. Add the exponents by putting both x^ into the same denominator (in this case 2 on the bottom). For example: (x^2)((1/2)(x^-1/2) -> (x^4/2)((1/2)(x^-1/2)) it becomes (1/2)(x^[4/2-1/2] => (1/2)(x^3/2) Its much easier to understand if you write it on paper! Link to post Share on other sites

iNow 6017 Posted May 16, 2009 Share Posted May 16, 2009 Its a bit hard to show using text, but its just basic exponent multiplication. Try using LaTex. Your text will look like this instead: [math](x^2-3x+2)({1}/{2}x^{-1/2})+(2x-3)(\sqrt{x})[/math] Link to post Share on other sites

paulo1913 10 Posted May 17, 2009 Author Share Posted May 17, 2009 I really still don't understand, mainly due to the set out of the explaination... I apreciate the effort but I can't make sense of it. Link to post Share on other sites

feign_ignorence 10 Posted May 17, 2009 Share Posted May 17, 2009 [math](x^2-3x+2)(\tfrac{1}{2}x^\frac{-1}{2})+(2x-3)(x^\frac{1}{2})[/math][math] (x^\frac{4}{2}-3x^\frac{2}{2}+2)(\tfrac{1}{2}x^\frac{-1}{2})+(2x^\frac{-6}{2}-3)(x^\frac{1}{2}) [/math] Same base = add the exponents. Add the exponents by putting both x^ into the same denominator (in this case 2 on the bottom). For example:[math] (x^2)(\tfrac{1}{2}x^\frac{-1}{2}) -> (x^\frac{4}{2})(\tfrac{1}{2}x^\frac{-1}{2})[/math] it becomes [math](\tfrac{1}{2}x^(\tfrac{4}{2}-\tfrac{1}{2}) => \tfrac{1}{2}x^\frac{3}{2}[/math] Its much easier to understand if you write it on paper! Thanks for that link for LaTeX Lets try this again! I hope my explaination makes a bit more sense now... Link to post Share on other sites

paulo1913 10 Posted May 18, 2009 Author Share Posted May 18, 2009 So can you show me how to get the final answer to this problem? I understand what you mean but I still don't get how to apply it properly Link to post Share on other sites

feign_ignorence 10 Posted May 19, 2009 Share Posted May 19, 2009 [math] (x^2-3x+2)(\tfrac{1}{2}x^\frac{-1}{2})+(2x-3)(x^\frac{1}{2}) [/math] [math] (x^\frac{4}{2}-3x^\frac{2}{2}+2)(\tfrac{1}{2}x^\frac{-1}{2})+(2x^\frac{2}{2}-3)(x^\frac{1}{2}) [/math] Multiply the terms through [math] (\tfrac{1}{2}x^\frac{3}{2}-\tfrac{3}{2}x^\frac{1}{2}+\tfrac{2}{2}x^\frac{-1}{2})+(2x^\frac{3}{2}-3x^\frac{1}{2}) [/math] Group terms with the same exponent ie:[math]x^\frac{1}{2}[/math] [math] (\tfrac{1}{2}x^\frac{3}{2}+2x^\frac{3}{2})+(-\tfrac{3}{2}x^\frac{1}{2}-3x^\frac{1}{2})+x^\frac{-1}{2} [/math] Only have one of each 'type' of x [math] (\tfrac{1}{2}x^\frac{3}{2}+\tfrac {4}{2}x^\frac{3}{2})+(-\tfrac{3}{2}x^\frac{1}{2}-\tfrac{6}{2}x^\frac{1}{2})+x^\frac{-1}{2} [/math] [math] (\tfrac{5}{2}x^\frac{3}{2}-\tfrac{9}{2}x^\frac{1}{2}+x^\frac{-1}{2}) [/math] Reduced form: [math] \tfrac{1}{2}x^\frac{-1}{2}(5x^\frac{4}{2}-9x^\frac{2}{2}+2) [/math] [math] \tfrac{1}{2}x^\frac{-1}{2}(5x^2-9x+2) [/math] (done here) If that ended up to be -2 instead: [math] \tfrac{1}{2}x^\frac{-1}{2}(5x^2-9x-2) [/math] It can be factored like this: [math] \tfrac{1}{2}x^\frac{-1}{2}(5x+1)(x-2) [/math] I think that's right. Link to post Share on other sites

the tree 222 Posted May 19, 2009 Share Posted May 19, 2009 Yeah, I got the same. Although I'd say it'd be easier to expand out before differentiating, and avoiding the chain rule altogether. Link to post Share on other sites

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