K M.

asked • 12/04/17

Differentiate the equation x^4y+xy^2=14 find the equation of the tangent line at the point (-2,1)

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Andy C. answered • 12/04/17

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Implicit differentiation featuring the product rule:
 
4x^3* y + x^4*y' + 2xyy' + y^2 = 0
 
(-2,1) --> x = -2 and y = 1
 
4(-2)^3*(1) + (1)^4*y' + 2(-2)(1)*y' + (1)^2 = 0
 
-32 + y' + -4y' + 1 = 0
 
-3y' - 31 = 0
 
y' = -31/3 = dy/dx = M
 
The tangent line:
 y = Mx + b
 
1 = (-31/3)*(-2) + b
1 = 62/3 + b
b = -59/3
 
y = (-31/3)X - 59/3
 
 
 
 
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Andrew M.

4(-2)^3*(1) + (1)^4*y' + 2(-2)(1)*y' + (1)^2 = 0
 
Andy, in the above line you accidentally plugged 1 in for x instead of -2...
That should be:
 
4(-2)^3*(1) + (-2)^4*y' + 2(-2)(1)*y' +(1)^2
-32 + 16y' - 4y' + 1
12y' - 31 = 0
y' = 31/12 = dy/dx = M
 
y = Mx + b
1 = (31/12)(-2) + b
1 = -31/6 + b
31/6 + 1 = b
b = 37/6
 
y = (31/12)x + 37/6
or, you can clear fractions by multiplying through by 12
12y = 31x + 74
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12/05/17

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