Final answer:
The slope of the tangent line to the curve y^{4}+2x^{2}y^{2}+5yx^{4}−5y=3 at the point (1,1), with y as the independent variable and x as the dependent variable, is determined by first computing the derivative of the equation with respect to y, and then evaluating this derivative at the point (1,1). This results in a slope of 8.
Step-by-step explanation:
To find the slope of the tangent line to the curve y^{4}+2x^{2}y^{2}+5yx^{4}−5y=3 at the point (1,1), with y as the independent variable and x as the dependent variable, we first need to take the derivative of the equation with respect to y. This gives us 4y^{3}+4xy^{2}+5x^{4}-5. Evaluating this derivative at the point (1,1) will provide the slope of the tangent line at that point, as the derivative represents the rate of change of the function at a given point. Hence, the slope of the tangent line at (1,1) is 4*1^{3}+4*1*1^{2}+5*1^{4}-5, which simplifies to 8.
The slope of a tangent line is equivalent to the derivative of the function at a given point, which signifies how fast the dependent variable (x) changes with respect to the independent variable (y) at that point. In the context of a graph, the slope indicates the steepness of the line at a particular point.
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