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Implicit differentiation serves as a technique used when dealing with equations that define one variable, typically 'y', implicitly in terms of another variable, like 'x'. In calculus, when you encounter a complex equation where 'y' cannot be easily isolated, implicit differentiation allows you to find the derivative \(\frac{dy}{dx}\) without explicitly solving for 'y'.

Here's a step-by-step guide to implement this method:

• Differentiate both sides of the equation with respect to 'x', treating 'y' as a function of 'x'.
• Remember that when you differentiate 'y' with respect to 'x', you apply the chain rule, which adds \(\frac{dy}{dx}\) to the derivative.
• Rearrange the equation to solve for \(\frac{dy}{dx}\), which gives you the slope of the tangent line at any point on the curve.

Using implicit differentiation makes finding the slope at a specific point, like \( (x_0, y_0) = (1,1) \), straightforward. Simply substitute the 'x' and 'y' values into the derived formula.

The slope of a tangent line to a curve at a given point is simply the value of the derivative of that curve at that point. The slope is a measure of how steep the line is at the point of tangency.

To find it, take these steps:

• Calculate \(\frac{dy}{dx}\) using the appropriate method, such as implicit differentiation if needed.
• Evaluate the derivative at the given point to ascertain the slope.
• If \(\frac{dy}{dx}\) at the point \( (x_0,y_0)\) is \(m\), then the slope of the tangent line at that point is 'm'.

In the exercise example, after using implicit differentiation, evaluating the derivative at \( (1,1)\) provides us with \(\frac{9}{11}\), the slope of the tangent line at that point.

The point-slope form is a format used to write the equation of a line when you are given a point on the line \( (x_0, y_0)\) and the slope of the line 'm'. The formula is expressed as:
\[y - y_0 = m(x - x_0)\]
It's a pivotal tool in calculus for converting the slope of the line and a reference point into an equation.

For instance, using the point-slope form with \( (x_0,y_0) = (1,1)\) and the slope \(m=\frac{9}{11}\), as we found from the previous concept, we can express the line equation as: \[y - 1 = \frac{9}{11}(x - 1)\]. This equation represents the tangent line that just 'grazes' the curve at the point \( (1,1)\).

A normal line to a curve at a given point is a straight line that is perpendicular to the tangent line at that point. The slope of the normal line is the negative reciprocal of the slope of the tangent line, which can be expressed as \(m_{normal} = -\frac{1}{m_{tangent}}\).

For example, if the slope of the tangent line is \(\frac{9}{11}\), then the slope of the normal line at that point would be \(m_{normal} = -\frac{11}{9}\). With this slope, and again using the point-slope form with the same point \( (x_0, y_0) = (1,1)\), the equation of the normal line becomes: \[y - 1 = -\frac{11}{9}(x - 1)\].

The normal line and tangent line equations provide two perpendicular lines that intersect at the curve, giving a clear visual of how the curve behaves at the tangent point.