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Chapter 5: Problem 78

Use implicit differentiation to find \(d y / d x\). $$ \ln x y+5 x=30 $$

Short Answer

Expert verified
The derivative of \(y\) with respect to \(x\), \(dy/dx\), given the equation \(\ln(x y) + 5x =30\) is \(-\frac{(y+x(1/y))}{x}\)

Step by step solution

01

Differentiate Both Sides of the Equation

The first step is to differentiate both sides of the equation \(\ln(x y) + 5x = 30\) with respect to \(x\). It's important to remember that \(\frac{d}{dx}\ln(f) = \frac{f'}{f}\), where \(f = x y\) in this case. Also, the derivative of a constant is zero.
02

Solve for \(dy/dx\)

After differentiation, rearrange the equation to isolate \(dy/dx\) on one side. This involves moving terms around and careful algebraic manipulation.
03

Simplify the Equation

Simplify the expression for \(dy/dx\) as far as possible by canceling common terms and combining like terms.

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Key Concepts

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

Logarithmic Differentiation
Logarithmic differentiation is a technique used to differentiate functions that are products, quotients, or powers of other functions. In this method, taking the natural logarithm of both sides allows for easier handling of complex products or quotients.

When faced with differentiating a function like \( x y\), we apply logarithmic rules to separate the terms:
  • Use the property that \((a \, b) = a + b\) to break down the expression \(\ln(xy) = x + y\).
  • This simplifies differentiation since we're now adding derivatives rather than multiplying or dividing.
This approach is particularly useful when the function involves variables multiplied together, as it turns a product into a sum, simplifying differentiation.
This technique proves powerful in solving equations that seem complex at first glance.
Partial Derivatives
Partial derivatives involve taking the derivative of a function with several variables with respect to one variable, treating all other variables as constants. In the context of our exercise, consider the function \((xy)\).

When differentiating \( (xy)\) with respect to \( x\), treat \( y\) as a constant:
  • Apply the product rule: if \(u = x\) and \(v = y\), then \( rac{d}{dx}(xy) = x rac{dy}{dx} + y\).
  • This helps isolate terms built around the variable of interest.
The concept of partial derivatives is essential when dealing with multi-variable functions, allowing you to isolate influence on one variable at a time.
This skill is fundamental when approaching problems that involve differentiating implicitly, as it requires careful separation of variables.
Chain Rule
The chain rule is an essential tool for differentiating compositions of functions. It states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function.

In the exercise, we encounter a composite function within \((xy)\):
  • Treat \( x \, y\) as a single function, requiring differentiation based on \( x\).
  • Apply the chain rule: differentiate the outer function \(\ln(u)\) where \(u = xy\), giving \( \frac{1}{xy}\), then multiply by the derivative of \(xy\).
  • This yields \(\frac{y + x \frac{dy}{dx}}{xy}\).
The chain rule's importance is in managing complex function layers, ensuring each step is correctly handled. This method enables you to navigate nested functions effectively.

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Most popular questions from this chapter

Linear and Quadratic Approximations Use a computer algebra system to find the linear approximation \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and the quadratic approximation \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) of the function \(f\) at \(x=a .\) Use a graphing utility to graph the function and its linear and quadratic approximations. $$ f(x)=\cosh x, \quad a=0 $$

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