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March 6th, 2017, 06:25 PM  #1 
Newbie Joined: Mar 2016 From: Canada Posts: 24 Thanks: 0  Finding points within a multivariable 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. 
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: $\displaystyle f(x,y)=(y+6)\ln(x)xe^{2y}x(y5)^5$ And so: $\displaystyle f_x(x,y)=(y+6)\cdot\frac{1}{x}e^{2y}(y5)^5$ $\displaystyle f_y(x,y)=\ln(x)2xe^{2y}5x(y5)^4$ Can you continue? 
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 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  

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calculus, finding, function, multivariable, points 
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