Question

In Exercises, find \(d y / d x\) implicitly. $$ 4 x^{3}+\ln y^{2}+2 y=2 x $$

Short answer

\(dy/dx = (2 - 12x^2) / (2/y + 2)\)

Step-by-step solution

Differentiate each term

Firstly, differentiate each term. For the first term, apply the power rule for differentiation, which states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). In the case of \(4x^3\), it differentiates to \(12x^2\). Note, \(dy/dx\) refers to the derivative of \(y\) with respect to \(x\). For \(\ln(y^2)\), the derivative of \(\ln(u)\) is \(1/u\), so multiply by the derivative of the inner function using the chain rule, resulting in \(2y/y^2\) or \(2/y\). Finally, \(2y\) differentiates using the chain rule to \(2 dy/dx\). Thus, the derivative of \(4x^3 + \ln(y^2) + 2y\) with respect to \(x\) is \(12x^2 + 2/y dy/dx + 2 dy/dx\)

Differentiate the other side of the equation

The derivative of \(2x\) with respect to \(x\) is simply \(2\).

Substitute equations

Substitute \(12x^2 + 2/y dy/dx + 2 dy/dx\) for \(d/dx(4x^3 + \ln(y^2) + 2y)\) and \(2\) for \(d/dx(2x)\) in the original equation \(4x^3 + \ln(y^2) + 2y = 2x\), the derivative of which is \(d/dx(4x^3 + \ln(y^2) + 2y) = d/dx(2x)\). So we get \(12x^2 + 2/y dy/dx + 2 dy/dx = 2\)

Solve for \(dy/dx\)

The goal is to solve for \(dy/dx\). This can be achieved by moving all terms not involving \(dy/dx\) to one side of the equation, to yield \(2/y dy/dx + 2 dy/dx = 2 - 12x^2\). To make \(dy/dx\) easier to isolate, factor it out to get \((2/y + 2) dy/dx = 2 - 12x^2\), and then divide by \((2/y + 2)\) to isolate \(dy/dx\), thus \(dy/dx = (2 - 12x^2) / (2/y + 2)\).

Key concepts

Chain Rule

When dealing with implicit differentiation, the chain rule is your best friend. It helps us differentiate composite functions, which are functions within functions. If you see a term like \(\ln(y^2)\), notice that it's not just \(y^2\) differentiated as a standalone piece. Here’s how the chain rule works:

• Differentiate the outer function, keeping the inside unchanged.
• Multiply by the derivative of the inner function.

For \(\ln(y^2)\), the outer function is the natural logarithm, and its derivative is \(1/u\), where \(u=y^2\). The inner function \(y^2\) differentiates to \(2y \cdot dy/dx\). Multiply these together to get \((1/y^2) \cdot 2y \cdot dy/dx = 2/y \cdot dy/dx\). This helps us differentiate terms with a variable not explicitly solved for.

Derivative

A derivative tells us how a function changes as its input changes. It’s the mathematical way of capturing the idea of rate of change.
When we perform implicit differentiation, we apply the derivative rules to each term but respect the relationships between variables that aren’t isolated. This means:

• Every time you differentiate \(y\) terms, include \(dy/dx\).
• Follow regular differentiation rules for terms with just \(x\).

For example, differentiating \(4x^3\) gives us \(12x^2\), which is straightforward as there is no \(y\) involved. Implicit differentiation comes in handy for equations where \(x\) and \(y\) are tangled up together.

Power Rule

The power rule is a basic, yet powerful tool in calculus. It's simple: if you have a function \(f(x) = x^n\), the derivative \(f'(x)\) is \(nx^{n-1}\).
In practice, this means:

• Drop down the exponent as a coefficient.
• Reduce the original exponent by one.

Consider \(4x^3\). Applying the power rule, we get \(3 \cdot 4x^{3-1} = 12x^2\). Easy, right? This rule applies only when the base is the variable you are differentiating with respect to, so apply it directly to terms like \(x\), but not to those involving \(y\) implicitly.

Natural Logarithm

The natural logarithm, represented by \(\ln(x)\), is a function that grows very slowly. It is the inverse of the exponential function \(e^x\). When differentiating \(\ln(u)\), where \(u\) is a function of \(x\), it’s essential to use the chain rule.

• The derivative of \(\ln(u)\) is \(1/u\).
• Multiply by the derivative of \(u\) itself.

For \(\ln(y^2)\), differentiate the inside \(y^2\) to get \(2y \cdot dy/dx\), and combine it with \(1/y^2\) from the derivative of \(\ln\). This results in \(2/y \cdot dy/dx\), linking the behavior of \(y\) in terms of \(x\). The natural logarithm’s derivative is a frequent guest in implicit differentiation problems due to its presence in composite functions.