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Asked Nov 5, 2019
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1. Use implicit differentiation to find an equation of the tangent line to the plot of a curve defined
by the relationship
T
6
dy
d
at the point (x, y) = (3, 7). You must justify all of your work in finding
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1. Use implicit differentiation to find an equation of the tangent line to the plot of a curve defined by the relationship T 6 dy d at the point (x, y) = (3, 7). You must justify all of your work in finding

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Expert Answer

Step 1

Given function and point is

- sin(x + y) = x -y
3 6
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- sin(x + y) = x -y 3 6

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Step 2

Using formula

Equation of line having slope m and passing through point (x, y)
(у - у) %3D т(х — х)
and also
d
d
dx
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Equation of line having slope m and passing through point (x, y) (у - у) %3D т(х — х) and also d d dx

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Step 3

First find the slope of the tangent line to the curve at given poi...

-sin(x + y)
dx6
dy
ла
{sin(x + y)} =1-
6 dx
dx
dy
d
-(x+ y) = 1 -
dx
-cos(x + y
6
dx
dy
=1--
dy
cos(x y1
dx
dy
dy
=1
dx
cos(x y)sx+ y):
6
dc
6
dy T
dx6
cos(x y
os(x+ y)+1} =1-
1-Tcos(+y)
dy
6
dx
cos(x+y)
is
3 6
Now, slope (m) at point
1-
cos
6
dy
6
п
cos
3
п
+1
6
6
1 -
cos
dy
+1
dy
1
л п
is
Now, slope (m) at point
1-
cos
6
dy
6
+
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-sin(x + y) dx6 dy ла {sin(x + y)} =1- 6 dx dx dy d -(x+ y) = 1 - dx -cos(x + y 6 dx dy =1-- dy cos(x y1 dx dy dy =1 dx cos(x y)sx+ y): 6 dc 6 dy T dx6 cos(x y os(x+ y)+1} =1- 1-Tcos(+y) dy 6 dx cos(x+y) is 3 6 Now, slope (m) at point 1- cos 6 dy 6 п cos 3 п +1 6 6 1 - cos dy +1 dy 1 л п is Now, slope (m) at point 1- cos 6 dy 6 +

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