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Implicit differentiation is a technique used in calculus. It lets us find derivatives of functions that are not easy to isolate. For example, we might have an equation like \(x^{4} y^{2} = 144\), where y is not by itself.
Usually, we differentiate each side of the equation with respect to x. But because y is a function of x, we must use the chain rule. In this method:

• First, differentiate both sides of the equation normally.
• Then, whenever you differentiate a term involving y, multiply by \(\frac{dy}{dx}\).

For the equation \(x^{4} y^{2} = 144\), differentiating implicitly with respect to x, we get: \[4x^{3} y^{2} + x^{4} \cdot 2y \cdot \frac{dy}{dx} = 0\]This technique is crucial for working with equations involving multiple variables.

Calculus is a branch of mathematics focusing on rates of change and the accumulation of quantities. It has two main parts: differential calculus and integral calculus.
Differential calculus deals with the concept of a derivative. This measures how a function changes as its input changes. In simpler terms, it's about understanding slopes of curves. For the curve given by \(x^{4} y^{2} = 144\), we used implicit differentiation to find the derivative or slope, \(\frac{dy}{dx}\).
Once we know the slope, we can find the tangent line to the curve at any given point. The tangent line touches the curve at just one point and has the same slope as the curve at that point.

A tangent line to a curve at a given point is a straight line that just 'touches' the curve at that point. The slope of this line is equal to the slope of the curve at that point.
To find the equation of the tangent line:

• Determine the slope of the curve at the point using derivative \(\frac{dy}{dx}\)
• Use the point-slope form of a line's equation: \[ y - y_0 = m(x - x_0)\]

Given the equation \(x^{4} y^{2} = 144\) and points (2,3) and (2,-3), implicit differentiation gives us slopes -1.5 and 1.5 respectively at these points.
For point (2,3), the tangent line's equation becomes: \(y - 3 = -1.5(x - 2)\), simplifying to \(y = -1.5x + 6\).
For point (2,-3), it becomes: \(y + 3 = 1.5(x - 2)\), simplifying to \(y = 1.5x - 6\).
This method involves understanding slopes and how they relate to the tangent lines touching the curves at specific points.