Asked by lena
what is the slope of the line tangent to the curve y^3+x^2y^2-3x^3=9 at the point (1,2)?
Answers
Answered by
victoria
Differentiating implicitly you get
3y^2y' + 2xy^2 + 2x^2yy' - 9x^2 = 0.
Solve for y', and substitute 1 for x and 2 for y to get the slope.
3y^2y' + 2xy^2 + 2x^2yy' - 9x^2 = 0.
Solve for y', and substitute 1 for x and 2 for y to get the slope.
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