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 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.

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