The coordinates of the point on the curve (x^2+1)(y-3)=x where a tangent to the curve has the gr

step 1 In this problem, solving for y from the given equation, we have y is equals to 3 plus x upon x s

The coordinates of the point on the curve (x^2+1)(y-3)=x...

The coordinates of the point on the curve $\left(x^{2}+1\right)(y-3)=x$ where a tangent to the curve has the greatest slope are given by
(a) $(\sqrt{3}, 3+\sqrt{3} / 4)$
(b) $(-\sqrt{3}, 3-\sqrt{3} / 4)$
(c) $(0,3)$
(d) $(\sqrt{3}, \sqrt{3} / 4)$

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Key Concepts

Tangent Line and Slope

The tangent line to a curve at a particular point is a straight line that just touches the curve at that point and has the same slope as the curve. The slope of the tangent line, given by the derivative dy/dx, indicates the rate at which y changes with respect to x at that specific point.

Optimization in Calculus

Optimization in calculus involves finding the maximum or minimum values of a function. When applied to a derivative, this can help identify points where the rate of change (in this case, the slope of the tangent line) is at its greatest or least. This is typically achieved by taking the derivative of the derivative (or using other optimization techniques) and setting it to zero to find critical points.

Implicit Differentiation

Implicit differentiation is the process used to find the derivative of a function that is given in an implicit form, where y is not isolated on one side of the equation. This technique involves differentiating both sides of the equation with respect to x while treating y as a function of x and applying the chain rule as necessary.

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