91Ó°ÊÓ

When dealing with equations involving two variables where one variable isn't isolated (like in our exercise), implicit differentiation is a crucial technique. It allows us to differentiate both sides of an equation with respect to a specific variable, often x, even if the equation is not explicitly solved for one variable. Differentiating implicitly comes in handy when equations have intricate combinations of x and y that are difficult to disentangle.

To perform implicit differentiation:

• Differentiate both sides of the equation with respect to x.
• Remember that when differentiating terms involving y, apply the chain rule by multiplying by \( \frac{dy}{dx} \) to account for y being a function of x.

In the exercise, the equation \(x^{2/3} + y^{2/3} = 5\) was differentiated implicitly, resulting in the expression \(\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0\). This requires solving for \(\frac{dy}{dx}\), which represents the slope of the tangent line at any point (x,y) on the curve.

The chain rule is an essential tool in calculus, especially when it comes to differentiating composite functions. It tells us how to take the derivative of a function that is composed of one or more other functions.

The basic idea is:

• To differentiate a composite function, find the derivative of the outer function first.
• Then multiply this by the derivative of the inner function.

In our context, when we differentiated \(y^{2/3}\) with respect to x, we had to consider y as a function of x. The expression \(\frac{d}{dx} y^{2/3}\) becomes \(\frac{2}{3}y^{-1/3} \times \frac{dy}{dx}\). This is a perfect example of where the chain rule steps in, reflecting the composition of functions: y raised to the power of \(2/3\) and then differentiated with respect to x.

Once we have the slope of a tangent line \(\frac{dy}{dx}\) from implicit differentiation, the next step is to write the equation of the tangent line. This is where the point-slope form of a line comes into play. It's particularly convenient when you know a point on the line and the slope.

The point-slope form is written as:

• \(y - y_1 = m(x - x_1)\)

Where \(m\) is the slope and \((x_1, y_1)\) is the known point on the line. From the exercise, the point given is \((8, 1)\), and after substituting into the derivative to find \(m\), we plug these into the point-slope form. This simplifies to give us the equation of the tangent line at the specified point on the curve.

Remember, the point-slope form is an easy way to express a line, bridging the connection between the analytical slope calculated and the graphical representation of the tangent line.