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 Pre-Calculus Pre-Calculus Math Forum

 April 14th, 2015, 08:50 AM #1 Newbie   Joined: Feb 2015 From: Scotland Posts: 20 Thanks: 0 Differentiation Hi All, I'm a bit stuck with this and any help would be appreciated... h(x) = sqrt(x^2-5) Determine the coordinates of the points where the tangent to the curve h(x) has gradient -3/2 I differentiated and got x/sqrt(x^2-5). First of all, is this correct? Second, if this x/sqrt(x^2-5) = -3/2 how do you solve it? I always have trouble with x when it's in a fraction. I tried x= -3(sqrt(x^2-5))/2 but got stuck again with the half power on (x^2-5). April 14th, 2015, 08:55 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 your derivative is correct $\displaystyle \frac{x}{\sqrt{x^2-5}} = -\frac{3}{2} \implies x < 0$ $\displaystyle 2x = -3\sqrt{x^2-5}$ $\displaystyle 4x^2 = 9(x^2-5)$ $\displaystyle x = -3$ Thanks from MathDino April 14th, 2015, 08:58 AM #3 Newbie   Joined: Feb 2015 From: Scotland Posts: 20 Thanks: 0 Wow, thanks. You make it look so easy. I had times it by two but missed squaring everything. Tags differentiation 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 Mr Davis 97 Calculus 1 March 9th, 2015 04:50 AM Soulanoid Calculus 1 September 29th, 2014 08:56 PM Ter Calculus 8 August 16th, 2012 09:12 AM florian101 Calculus 1 December 22nd, 2009 06:42 AM MyNameIsVu Calculus 1 June 3rd, 2009 05:18 AM

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