Question

Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=10 at the point (5,1) . The equation of this tangent line can be written in the form y=mx+b
where m is:
and b is:

Answer 1

first step is to find the derivative of \(y\) wrt \(x\)
start with
\[y^3+3x^2y'+y+xy'=0\] replace \(x\) by \(5\) and \(y\) by \(1\) and solve for \(y'\)

Answer 2

y prime= -10?

Answer 3

y^3 + 3y^2* y' *x + y + xy' = 0

y' = -(y^3 +y)/(x+3y^2 *x)

Answer 4

-1/10

Answer 5

and then whats next?

Answer 6

that's your 'm' (for the tangent line)
use point-slope or
y=mx+b

1=(-1/10) *5 +b
b=...

Answer 7

3/2 ?

Answer 8

yes

Answer 9

all good? questions?

Answer 10

i understand i got 3/8 but i understand what i did wrong thanks!