My Math Forum Tangent line equation

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 May 25th, 2012, 03:31 PM #1 Member   Joined: Jan 2012 Posts: 43 Thanks: 0 Tangent line equation Hey, I'm having trouble with a quotient rule problem that asks you to find the equation of the tangent line. The function is $\frac{sqrt(x)*(3-2x^2)}{x}$ The question then states that when x=9 the corresponding y value will be______ and the slope of the tangent line Is f'(x)=9=_______. Thererore the equation of the tangent line is_______? In form ax+b. For the first blank I just inserted x=9 into the original equation and solved f(x)=-53. For the second blank I got the slope equalling -55/9 by plugging x=9 into the derivative of f(x). I Calculated the Derivative using the product rule $sqrt(x)\frac{x(-2)-(3-2x^2)(1)}{x^2}$ And ended up with a derivative of: $\frac{-(2x^2+3)sqrt(x)}{x^2}$ For the tangent line equation I got $y=\frac{-55}{9}x-2$ By plugging in the coordinates (9,-53) to y+53=(-55/9)x-9 Some part here is incorrect I am not sure which, any help would be much appreciated
 May 25th, 2012, 03:39 PM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Tangent line equation If you plug x=9 into the derivative $f'(x)=\frac{-3(2x^{2}+1)}{2x^{3/2}}$, you should get $f'(9)=\frac{-163}{18}$. Also, $f(9)=-53$, as you correctly have. So, you now have x=9, y=-53, m=-163/18. All set to find line equation. It may be a little easier to find the derivative by expanding f(x) into $3x^{\frac{-1}{2}}-2x^{\frac{3}{2}}$ Then, term-by-term, we get $f'(x)=-3\sqrt{x}-\frac{3}{2}x^{\frac{-3}{2}}$. Or you can write it as I have above. Just another equivalent form. Is that a picture of you with a TI?. Good to see you're into math. Count yourself as a member of a small minority of the overall population.
 May 25th, 2012, 03:59 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Tangent line equation Though galactus mentioned it (while I was posting ), here is another approach: $f(x)\,=\,\frac{\sqrt{x}(3\,-\,2x^2)}{x}\,=\,\frac{3\sqrt{x}}{x}\,-\,\frac{\sqrt{x}2x^2}{x}\,=\,3x^{-1/2}\,-\,2x^{3/2}$ Now it is a little easier to differentiate. $f'(x)\,=\,-\frac32 x^{-3/2}\,-\,3x^{1/2},\,f#39;(9)\,=\,-\frac{163}{18}$
 May 25th, 2012, 10:32 PM #4 Member   Joined: Jan 2012 Posts: 43 Thanks: 0 Re: Tangent line equation Ya thanks guys I haven't dealt with product quotient rules in awhile so the different approaches are very helpful as i get back into things. And ya galactus I've always enjoyed math, although the program I'm in is calling for less and less of it as I near the end, unfortunately..

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