Calc2.45

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Find the slope of the tangent line to the graph xy=y^{2}x^{2}+y+x+2 at the point (0,-2).

Again, use implicit differentiation to find y' and then plug in the point (0,-2) to find the slope of the tangent line at that point. Taking the derivative gives

y+xy'=2yy'x^{2}+y^{2}(2x)+y'+1\,

Plugging in (0,-2) gives

-2+0=0+0+y'+1\,

or

y'=-3\,

Thus, the slope of the tangent line to this graph at the point (0,-2) is -3. Since we have a point and the slope of the tangent line, we will use the point-slope form of the line. The equation of the tangent line at (0,-2) is

y-(-2)=-3(x-0)\,

or

y+2=-3x\,


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