Problem 12 - Calculus - AP Edition

Question

Find \(d y / d x\) by implicit differentiation. $$\frac{x y^{3}}{1+\sec y}=1+y^{4}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \text{computed expression from the steps.}\)

Step 1: Differentiate Both Sides

Differentiate each side of the equation with respect to \(x\). Remember that \(y\) is a function of \(x\), so use the chain rule when differentiating terms involving \(y\).

Step 2: Differentiate Left Side Using Product and Chain Rule

For the left side, \(\frac{xy^3}{1+\sec y}\), first apply the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \) where \(u = xy^3\) and \(v = 1+\sec y\). Find \(u'\) and \(v'\) using the product rule and chain rule respectively.

Step 3: Differentiate Right Side

For \(1 + y^4\), differentiate to get \(0 + 4y^3 \frac{dy}{dx}\). This is straightforward application of basic differentiation and the chain rule.

Step 4: Substitute and Simplify Derivatives

From steps 2 and 3, substitute the derivatives back into the differentiated equation. Simplify the expression where possible.

Step 5: Isolate \(\frac{dy}{dx}\)

Collect all terms involving \(\frac{dy}{dx}\) on one side of the equation and solve for \(\frac{dy}{dx}\). This involves algebraic manipulation to isolate \(\frac{dy}{dx}\).

Key Concepts

Chain Rule

When working with implicit differentiation, the chain rule is a crucial technique. It allows us to differentiate composite functions, where one function is nested inside another. In our original exercise, since we are dealing with an equation combining both \(x\) and \(y\), and knowing that \(y\) is a function of \(x\), the chain rule comes into play to help us correctly find derivatives involving \(y\).
The chain rule states that if you have a function \(z = f(g(x))\), the derivative \(\frac{dz}{dx}\) is \(f'(g(x)) \cdot g'(x)\). This means we differentiate the outer function as if the inner function \(g(x)\) were just \(x\), then multiply by the derivative of the inner function itself. In the problem, you'll see this when differentiating terms like \(y^3\), giving us \(3y^2 \frac{dy}{dx}\).

• Identify inner and outer functions in your equation.
• Differentiate the outer function normally.
• Multiply the result by the derivative of the inner function.

Quotient Rule

The quotient rule is used for differentiating expressions that are ratios of two functions. In our equation, we encounter the term \(\frac{xy^3}{1+\sec y}\). To differentiate this term, we have to apply the quotient rule because it is of the form \(\frac{u}{v}\).
The quotient rule formula is given by:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
Here, \(u = xy^3\) and \(v = 1+\sec y\). You need to differentiate \(u\) and \(v\) individually. It's essential to additionally apply the product rule for \(u = xy^3\), as it is a product of two separate functions. Once you have \(u'\) and \(v'\), plug them into the quotient rule to get the derivative of the ratio.

• Differentiate numerator and denominator separately.
• Use the quotient rule formula to combine these derivatives.

Product Rule

The product rule helps differentiate expressions that are products of two functions. Within the \(u = xy^3\), as identified in the quotient rule, we observe a product of \(x\) and \(y^3\). Thus, the product rule needs to be employed to find the derivative of this term.
The product rule formula is:
\[ (fg)' = f'g + fg' \]
In our exercise:

• Let \(f = x\) and \(g = y^3\).
• This gives \(f' = 1\) and \(g' = 3y^2 \frac{dy}{dx}\).
• Using the product rule, \( (xy^3)' = 1 \cdot y^3 + x \cdot 3y^2\frac{dy}{dx} \).

Combine these to complete the differentiation of \(xy^3\), supplying \(u'\) required in the quotient rule.

Differentiation Techniques

Implicit differentiation brings several techniques together, creating a powerful toolkit for handling complex equations. When dealing with implicit functions like in our exercise, it is vital to strategically use various differentiation rules.

• Chain Rule: Required when differentiating composite functions, especially where one variable is nested within another as a function.
• Quotient Rule: Essential for fractions, allowing differentiation of ratios by applying its distinct formula.
• Product Rule: Necessary for terms which are products of straightforward functions, ensuring each component is properly differentiated.
• Simplification and Rearrangement: Post-differentiation, simplification and rearranging terms are vital to isolate \(\frac{dy}{dx}\), thus solving for the derivative demanded.

The combination of these techniques allows you to meticulously tackle and solve equations involving both \(x\) and \(y\) intertwined in complex forms we see in implicit differentiation challenges.