the question is find the equation of the tangent line to the curve 2x^3 + 2y^3 = 9xy when x=1, y=2 i tried d2x^3/dx + d2y^3/dx = d9xy/dx d2x^3/dx = 6x^2 d2y^3/dx = 6y^2 d9xy/dx = ? i don't know how to do how do i do this problem?
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Originally Posted by haebinpark the question is find the equation of the tangent line to the curve 2x^3 + 2y^3 = 9xy when x=1, y=2 i tried d2x^3/dx + d2y^3/dx = d9xy/dx d2x^3/dx = 6x^2 d2y^3/dx = 6y^2 d9xy/dx = ? i don't know how to do how do i do this problem? Use implicit differentiation. The RHS will require the product rule. Looking at the RHS: The LHS is simpler: Set both sides equal to each other and isolate dy/dx. You can then sub in the value you know to find the gradient of the tangent Spoiler: A factor of 3 will also cancel out:
but you have dy/dx on both side?
anyone...?
Originally Posted by e^(i*pi) Use implicit differentiation. The RHS will require the product rule. Looking at the RHS: The LHS is simpler: Set both sides equal to each other and isolate dy/dx. You can then sub in the value you know to find the gradient of the tangent Spoiler: A factor of 3 will also cancel out: can i go from 6x^2 + 6y^2 dy/dx = 9y+9x dy/dx to 6y^2 dy/dx - 9x dy/dx = 9y - 6x^2 ? and after finding dy/dx how do i find the equation? because that was the question... hmm
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