My Math Forum Find the equation of the tangent line to G(x) .

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 October 6th, 2013, 04:24 PM #1 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Find the equation of the tangent line to G(x) . Find the equation of the tangent line to $g(x)= arctan(x) + ln(x)$ at $x= 1$ so; $G(x) = arctan(x) + ln(x) G(1) = arctan(1) + ln(1) y = \frac{\pi}{4} + 0 y = \frac{\pi}{4}$ so I now have $x$ & $y$.. $(1, \frac{\pi}{4})$ so I take derivative to find slope: $G'(x) = (arctan(x))'(ln(x)) + (arctan(x))(ln(x))' G'(x) = \frac{1}{1 + x^2} * ln(x) + (arctan(x)) * \frac{1}{x} G'(x) = \frac{1}{1+ 1^2} * ln(1) + arctan(1) * \frac{1}{1} m = \frac{\pi}{4} * \frac{1}{1} = \frac{\pi}{4}$ so $m= \frac{\pi}{4}$ so $y - \frac{\pi}{4}= \frac{\pi}{4}(x - 1)$ ?
 October 6th, 2013, 05:08 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,806 Thanks: 1045 Math Focus: Elementary mathematics and beyond Re: Find the equation of the tangent line to G(x) . $g(x)\,=\,\arctan(x)\,+\,\ln(x)$ At x = 1, y = $\frac{\pi}{4}$ $g'(x)\,=\,\frac{1}{1\,+\,x^2}\,+\,\frac1x,\,g#39;(1)\,=\,\frac32$ Tangent line: $y\,-\,\frac{\pi}{4}\,=\,\frac32(x\,-\,1)$
 October 7th, 2013, 05:12 AM #3 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Re: Find the equation of the tangent line to G(x) . Oh wow, I see now. I was using the product rule when it was clearly unnecessary. So I take derivative then plug in a 1. Ok got it.

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