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March 6th, 2017, 06:25 PM   #1
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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
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 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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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.

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