91Ó°ÊÓ

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. When we have a function defined implicitly, such as in the given exercise, the chain rule becomes vital.

The chain rule states that if we have a composite function \(f(g(x))\), then its derivative can be found using:
\frac{d}{dx}[ f(g(x))] = f'(g(x)) \times g'(x).

In our exercise, we applied the chain rule to differentiate \(y^3\) with respect to \(x\). Since \(y\) is a function of \(x\), we treated \(y\) as \(g(x)\) and \(y^3\) as \(f(y)\). Here's how we apply the chain rule to \(4y^3\): Firstly, differentiate the outer function \(y^3\) to get \(3y^2\) and then multiply it by the derivative of the inner function \(y\) with respect to \(x\), denoting it as \(\frac{dy}{dx}\). Hence, the differentiation looks like this: \(4 \frac{d}{dx}(y^3) = 4 \times 3y^2 \frac{dy}{dx} = 12y^2 \frac{dy}{dx}\). Notice how we multiply by \(12\) since the constant multiplies throughout.

Remember to always apply the chain rule when differentiating composite functions!

Implicit differentiation is a technique used when a function is not given explicitly, but implicitly, in a relationship involving both \(x\) and \(y\).

In our given equation, \texttt{4y^3 - x^2 = -5}, \(y\) is defined implicitly in terms of \(x\). Direct differentiation is not possible because \(y\) is not isolated.

To proceed, differentiate every term with respect to \(x\). For left side: \( \frac{d}{dx}(4y^3) - \frac{d}{dx}(x^2)\) gives \( 4\frac{d}{dx}(y^3) - 2x\). And the differentiation of constant \( -5\) on the right side is zero. This leaves us with \(4\frac{d}{dx}(y^3) - 2x = 0\). Another layer of differentiation is required for \(y^3\) due to chain rule as it needs \(\frac{dy}{dx}\) while \(x\) differentiation is direct.

Hence, with the intermediate steps, we arrived at \(12y^2\frac{dy}{dx} - 2x = 0\). The implicit differentiation allows us to handle mixed \(x\) and \(y\) relations effectively!

Once we have differentiated the equation implicitly, our next step is to solve for the derivative, \(\frac{dy}{dx}\).

From step 2, the equation is \(12y^2\frac{dy}{dx} - 2x = 0\). To isolate \(\frac{dy}{dx}\), first move \(2x\) to the right side: \(12y^2\frac{dy}{dx} = 2x\). Then, divide both sides by \(12y^2\) to isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{2x}{12y^2} = \frac{x}{6y^2}\).

This gives us the general form of the slope at any point \((x,y)\) on the curve.

Finally, substitute the specific point \((x=3, y=1)\) provided in the exercise to find the slope at that point. Substitute \(x=3\) and \(y=1\) into \(\frac{dy}{dx} = \frac{x}{6y^2}\) to get:
\frac{dy}{dx} = \frac{3}{6(1)^2} = \frac{3}{6} = \frac{1}{2}.

Thus, the slope of the graph at the given point \((3, 1)\) is \(\frac{1}{2}\). This is the process for solving derivatives in implicit relationships: differentiate implicitly, isolate mixed derivative \(\frac{dy}{dx}\) and then substitute given point values.

Practice with different equations to solidify understanding!