use implicit differentiantion to find the equation of the tangent line to the curve xy^3+xy=20 at the point (10,1) . The equation of this tangent line can be written in the form y=mx+b where m is=
and where b is =
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First, solve for the slope of the tangent line. To do so, differentiate both sides of the equation with respect to x.
Since the derivative of constant is zero, right side becomes:
`d/(dx) (xy^3+xy)= 0`
At the left side, take the derivative of each term.
Then, apply product rule of derivative.
`y^3*1 + x*3y^2 dy/dx + y*1+x*dy/dx =0`
Move the terms without dy/dx to the other side of the equation.
Then, factor out dy/dx.
And, divide both sides by 3xy^2+x to get dy/dx.
Now that dy/dx is known, plug-in (10,1) to get the slope of the tangent line.
So, the slope of the line tangent to the curve at (10,1) is -1/20.
Next, apply the point-slope form to get the equation of the line.
And, add both sides by 1.
Hence, the equation of the line is `y=-x/20+3/2` . Its slope is `m=-1/20` and y-intercept is `b=3/2` .
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