Question

Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy= 14 at the point ( 7 , 1 ). The equation of this tangent line can be written in the form y = mx+b where m is:

and b is:

Answer 1

Answer:
`$m=-(1)/(22)$`
`$b=(15)/(22)$`

Explanation:

To find the equation of the tangent line to the curve
`$xy^3+xy= 14$` at the point
`$(7,1)$` we need to first find the derivative of the equation with respect to x using implicit differentiation.

Taking the derivative of both sides with respect to x:

`$(d)/(dx)(xy^3+xy)=(d)/(dx)(14)$`

Using the product rule and chain rule:

`$y^3+x(3y^2(dy)/(dx)+y)=0$`

Simplifying and solving for
`$(dy)/(dx)$`

`$(dy)/(dx)=-(y)/(3y^2+1)$`

Now we can substitute x=7 and y=1 into this equation to find the slope of the tangent line at the point (7,1):

`$m=\left.-(y)/(3y^2+1)\right|_((7,1))=-(1)/(22)$`

So the slope of the tangent line is
`$-(1)/(22)$`

To find the y-intercept, we can use the point-slope form of the equation of a line:

`$y-1 = m(x-7)$`

`$y=mx+(1-7m)$`

Therefore, the equation of the tangent line to the curve
`$xy^3+xy=14$` at the point (7,1) can be written as:

`$y=-(1)/(22)x+(15)/(22)$`

So
`$m=-(1)/(22)$` and
`$b=(15)/(22)$`