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March 6th, 2017, 06:25 PM   #1
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Thanks: 0 Finding points within a multi-variable calculus function

Consider the following function.

f (x, y) = [(y+6) ln x]−xe^2y−x(y−5)5f (x, y) = [(y+6) ln x]−xe^2y−x(y−5)5
(a) Find  fx(1, 0)fx(1, 0) .

(b) Find  fy(1, 0)fy(1, 0) .

I know I need to take the partial derivatives of ff and evaluate them at the same point. I added a pic of the question for verification. Please help me find the answer.
Attached Images q7.jpg (11.7 KB, 2 views) March 6th, 2017, 06:48 PM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs We are given: $\displaystyle f(x,y)=(y+6)\ln(x)-xe^{2y}-x(y-5)^5$ And so: $\displaystyle f_x(x,y)=(y+6)\cdot\frac{1}{x}-e^{2y}-(y-5)^5$ $\displaystyle f_y(x,y)=\ln(x)-2xe^{2y}-5x(y-5)^4$ Can you continue? Thanks from puppypower123 March 6th, 2017, 07:19 PM #3 Newbie   Joined: Mar 2016 From: Canada Posts: 24 Thanks: 0 So then do I plug in my points? so f_x = 3130 and f_y = -3127? because fx(x,y)=(0+6)⋅1(1)−e^2(0)−(y−5)^5 =3130 and fy(x,y) = ln(1)−2(1)e^2(0)−5(1)(0−5)^4 = -3127 Last edited by puppypower123; March 6th, 2017 at 07:22 PM. March 6th, 2017, 07:55 PM   #4
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From: St. Augustine, FL., U.S.A.'s oldest city

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Math Focus: Calculus/ODEs
Quote:
 Originally Posted by puppypower123 So then do I plug in my points? so f_x = 3130 and f_y = -3127? because fx(x,y)=(0+6)⋅1(1)−e^2(0)−(y−5)^5 =3130 and fy(x,y) = ln(1)−2(1)e^2(0)−5(1)(0−5)^4 = -3127
Yes, that's correct.  Tags calculus, finding, function, multivariable, points 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 JulieK Calculus 3 August 11th, 2014 12:59 AM irunktm Algebra 6 September 7th, 2012 12:42 PM Chasej Calculus 7 September 16th, 2011 12:18 AM person1200 Calculus 1 September 12th, 2010 07:17 PM wetmelon Algebra 2 March 23rd, 2009 11:00 PM

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