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Chapter 3: Problem 5

For each equation, use implicit differentiation to find \(d y / d x\). $$ y^{4}-x^{3}=2 x $$

Short Answer

Expert verified
\(\frac{dy}{dx} = \frac{2 + 3x^2}{4y^3}\).

Step by step solution

01

Differentiate Both Sides with Respect to x

The given equation is \(y^4 - x^3 = 2x\). To apply implicit differentiation, we need to differentiate both sides with respect to \(x\). This means taking the derivative of each term separately:- The derivative of \(y^4\) with respect to \(x\) is \(4y^3 \frac{dy}{dx}\). This requires using the chain rule.- The derivative of \(-x^3\) with respect to \(x\) is \(-3x^2\).- The derivative of \(2x\) with respect to \(x\) is \(2\). Thus, the differentiated equation is:\[4y^3 \frac{dy}{dx} - 3x^2 = 2.\]
02

Solve for dy/dx

Our task is to solve for \(\frac{dy}{dx}\). Rearrange the differentiated equation for this:\[4y^3 \frac{dy}{dx} = 2 + 3x^2.\]Now, solve for \(\frac{dy}{dx}\) by dividing both sides by \(4y^3\):\[\frac{dy}{dx} = \frac{2 + 3x^2}{4y^3}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Chain Rule
When we deal with implicit differentiation, the chain rule becomes a powerful tool. The chain rule helps us find the derivative of a composite function. For instance, if we have a function like \( y^4 \) where \( y \) itself is a function of \( x \), we'll use the chain rule to differentiate it.

The main idea of the chain rule is to differentiate the outer function and multiply by the derivative of the inner function. Here's how it applies:
  • Differentiate the outer function: For \( y^4 \), the outer function is \( u^4 \), so its derivative is \( 4u^3 \).
  • Multiply by the derivative of the inner function: Since \( u = y \) and \( y = y(x) \), we multiply by \( \frac{dy}{dx} \).
Thus, the derivative of \( y^4 \) with respect to \( x \) becomes \( 4y^3 \frac{dy}{dx} \). This allows us to work with more complex relationships between variables in calculus.
Exploring Derivatives
Derivatives are the fundamental building blocks of calculus. They measure how a function changes as its input changes. In simpler terms, the derivative tells us the rate of change or the slope of a function at any given point.

When performing implicit differentiation, we differentiate each part of the equation with respect to \( x \). Let's see how derivatives were applied in the original solution:
  • The derivative of \( y^4 \) with respect to \( x \) involves both the chain rule, making it \( 4y^3 \frac{dy}{dx} \).
  • For \( -x^3 \), the derivative with respect to \( x \) is simply \( -3x^2 \), using basic derivative rules.
  • Finally, for \( 2x \), the derivative with respect to \( x \) is \( 2 \).
Together, these derivatives transform the equation into a form that lets us easily solve for unknown derivatives like \( \frac{dy}{dx} \). Understanding these changes is crucial for mastering calculus.
Rearranging Equations Effectively
After differentiating both sides of an equation, our goal is to isolate \( \frac{dy}{dx} \). This process often involves rearranging equations to express one variable in terms of others. Let's break down how this is done.
  • Start with the differentiated equation: \( 4y^3 \frac{dy}{dx} - 3x^2 = 2 \).
  • First, move terms not involving \( \frac{dy}{dx} \) to the other side: add \( 3x^2 \) to both sides to get \( 4y^3 \frac{dy}{dx} = 2 + 3x^2 \).
  • Now, solve for \( \frac{dy}{dx} \) by dividing both sides by \( 4y^3 \). This isolates \( \frac{dy}{dx} \), giving us: \( \frac{dy}{dx} = \frac{2 + 3x^2}{4y^3} \).
Rearranging involves methodically shifting parts of the equation around, while maintaining balance. This strategy is key in finding derivatives implicitly and is widely used in algebra and calculus alike.

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Most popular questions from this chapter

$$ x^{2}+x y=y^{2} $$

For each equation, use implicit differentiation to find \(d y / d x\). $$ y^{3}-y^{2}+y-1=x $$

For each demand equation, use implicit differentiation to find \(d p / d x\). $$ 12 p^{2}+4 p+1=x $$

True or False: If a function is defined and continuous on a closed interval, then it has both an absolute maximum value and an absolute minimum value.

Find the equation of the tangent line to the curve at the given point using implicit differentiation. $$ \begin{array}{l} \text { Cardioid }\\\ \left(x^{2}+y^{2}-y\right)^{2}=x^{2}+y^{2} \quad \text { at } \quad(-1,0) \end{array} $$
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