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Math Help - the slope of the tangent to the curve (y^3)x+(y^2)x^6=6 at 2,1

  1. #1
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    the slope of the tangent to the curve (y^3)x+(y^2)x^6=6 at 2,1

    Using implicit differentiation, I get a ridiculous slope when I plug in the 2 and 1, m=-193/134.
    I get -y^3-6x^5y^2 all divided by 3y^2x+2yx^6.

    Can someone point out where my mistake is? I did use the product rule for xy...so not sure where I messed up.
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  2. #2
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    Quote Originally Posted by bcahmel View Post
    Using implicit differentiation, I get a ridiculous slope when I plug in the 2 and 1, m=-193/134.
    I get -y^3-6x^5y^2 all divided by 3y^2x+2yx^6.

    Can someone point out where my mistake is? I did use the product rule for xy...so not sure where I messed up.
    The first obvious problem is that the point (2,1) does not lie on the curve (y^3)x + (y^2)x^6 = 6.
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  3. #3
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    oh. so I take the derivative of the function to get the slope.Then y-2=derivative(x-1) for the line.
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    Quote Originally Posted by bcahmel View Post
    oh. so I take the derivative of the function to get the slope.Then y-2=derivative(x-1) for the line.
    The question asked for the slope of the tangent to the curve (y^3)x+(y^2)x^6=6 at (2,1). But if the point (2,1) does not lie on the curve then the question does not make any sense.

    I suspect that there is a mistake in the question, and that it ought to ask for the slope of the tangent to the curve (y^3)x+(y^2)x^2=6 at (2,1). If x^6 is replaced by x^2 in that way, then the point does lie on the curve. You can now use implicit differentiation as in the original post, and this time it should lead to a more reasonable answer.
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