Question
Asked Oct 7, 2019

Find the slope of the tangent line to the curve defined by

6x+ 8xy +y= -13 at (-1,1).

The slope of the curve at the point (-1,1) is ?

check_circleExpert Solution
Step 1

The slope of tangent line to the curve f(x,y)=0 at (a,b) is the value of dy/dx at (a,b)

Step 2

We differentiate both sides of given equation of curve to find dy/dx

6x8.xy y
-13
dy
dy
30x48
- 2y
dc
0 =
dx
dy 30x4-8y
(8х + 2у)
dx
dy 30x-8y
8x+2y
dx
help_outline

Image Transcriptionclose

6x8.xy y -13 dy dy 30x48 - 2y dc 0 = dx dy 30x4-8y (8х + 2у) dx dy 30x-8y 8x+2y dx

fullscreen
Step 3

Then we substitute x=-1 and y=1 in dy/dx to get the value o...

-30(-1)-8(1) 19
8(-1)+2(1)
dy
т
dх
(-1,1)
3
help_outline

Image Transcriptionclose

-30(-1)-8(1) 19 8(-1)+2(1) dy т dх (-1,1) 3

fullscreen

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Calculus

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