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

1. ## 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.

2. Originally Posted by bcahmel
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.

3. oh. so I take the derivative of the function to get the slope.Then y-2=derivative(x-1) for the line.

4. Originally Posted by bcahmel
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.