Find the equation of the tangent to the curve x+xy+y=5 at x=5 help?

2 Answers
Noah G
Jan 10, 2018

y=56−16x

Explanation:

Start by finding the y-value:

5+5y+y=5

6y=0

y=0

Now we find the derivative using implicit differentiation.

1+y+x(dydx)+dydx=0

y+x(dydx)+dydx=−1

x(dydx)+dydx=−1−y

dydx(x+1)=−1−y

dydx=−1−yx+1

dydx=−y+1x+1

At (5,0), the derivative will have value

dydx=−16

Now use point-slope form to find the equation:

y−y1=m(x−x1)

y−0=−16(x−5)

y=−16x+56

Hopefully this helps!

Douglas K.
Jan 10, 2018

y=−16x+56

Explanation:

Begin by finding the y coordinate at x=5:

5+5y+y=5

5y+y=0

6y=0

y=0

The point of tangency is (5,0)

Compute the first derivative of the curve:

1+y+xdydx+dydx=0

(x+1)dydx=−(y+1)

dydx=−y+1x+1

The slope, m, of the tangent line is the first derivative evaluated at the point (5,0):

m=−0+15+1

m=−16

Use the point-slope form for the equation of a line:

y=m(x−x0)+y0

y=−16(x−5)+0

y=−16x+56

Here is the a graph of the curve and the tangent line:

![www.desmos.com/calculator](useruploads.socratic.org)

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