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August 1, 2015

August 1, 2015

Posted by **DJ** on Friday, October 28, 2011 at 6:18pm.

at the point (1, -6/55)

Use implicit differentiation to find the slope of the tangent line to the curve

- Calculus -
**Reiny**, Friday, October 28, 2011 at 7:25pmcross-multiply first

y = x^3 - 7x + 9x^2y - 63y

y' = 3x^2 - 7 + 9x^2y' + 18xy - 63y' , where y' = dy/dx

64y' - 9x^2y' = 3x^2 + 18xy-7

y' = dy/dx = (3x^2 + 18xy - 7)/(64 - 9x^2)

I will leave it up to you to do the remaining arithmetic