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Chapter 7: Problem 38

Use implicit differentiation to find \(y^{\prime}\). $$ y^{3}+x^{2} \ln y=5 x+3 $$

Short Answer

Expert verified
\(y' = \frac{5 - 2x \ln y}{3y^2 + \frac{x^2}{y}}\).

Step by step solution

01

Differentiate Both Sides

Differentiate both sides of the equation with respect to \(x\). Remember that \(y\) is a function of \(x\), so you will need to apply the chain rule. The differentiation will look like this: \[ \frac{d}{dx}(y^3) + \frac{d}{dx}(x^2 \ln y) = \frac{d}{dx}(5x + 3). \]
02

Differentiate Terms

For the first term \(y^3\), use the chain rule: \[ \frac{d}{dx}(y^3) = 3y^2 \cdot y'. \] For the second term \(x^2 \ln y\), apply the product rule:\[ \frac{d}{dx}(x^2 \ln y) = \frac{d}{dx}(x^2) \cdot \ln y + x^2 \cdot \frac{d}{dx}(\ln y) = 2x \ln y + x^2 \cdot \frac{1}{y} \cdot y'. \]The third term, on the right side, is simplified as:\[ \frac{d}{dx}(5x + 3) = 5. \]
03

Combine the Differentiated Terms

Now substitute the differentiated terms back into the equation:\[ 3y^2 \cdot y' + 2x \ln y + x^2 \cdot \frac{1}{y} \cdot y' = 5. \]
04

Solve for \(y'\)

Combine the terms that contain \(y'\) to isolate it:\[ y'(3y^2 + \frac{x^2}{y}) = 5 - 2x \ln y. \]Finally, solve for \(y'\):\[ y' = \frac{5 - 2x \ln y}{3y^2 + \frac{x^2}{y}}. \]

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Key Concepts

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

Chain Rule
Understanding the chain rule is a vital part of implicit differentiation. This rule helps us differentiate compositions of functions smoothly. When we have a function like \(y = g(x)\), where \(y\) is a function of \(x\), using the chain rule allows us to differentiate expressions where \(y\) isn鈥檛 isolated. In our exercise, when differentiating \(y^3\), you apply the chain rule by taking the derivative of the outer function first and then the inner function.
  • Start by differentiating the outer part: \(y^3\), which becomes \(3y^2\).
  • Then multiply by the derivative of the inner function \(y\) with respect to \(x\), which is \(y'\).
So, for \(y^3\), it becomes \(3y^2 \cdot y'\). The chain rule systematically helps tackle complex problems where functions operate within other functions.
Product Rule
The product rule is the go-to rule for differentiating terms that are products of two functions. In our scenario, we handle the term \(x^2 \ln y\) using the product rule. The product rule states that for two functions \(f(x)\) and \(g(x)\), the derivative of their product is:\[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x). \]For \(x^2 \ln y\), look at it as a product of two functions: \(f(x) = x^2\) and \(g(x) = \ln y\).
  • Differentiating \(f(x) = x^2\) gives \(2x\).
  • Keep \(g(x)\) as it is: \(\ln y\).
  • Differentiating \(g(x) = \ln y\) gives \(\frac{1}{y}\cdot y'\), by the chain rule since \(y\) is a function of \(x\).
  • Combine them: \(2x \ln y + x^2 \cdot \frac{1}{y} \cdot y'\).
These steps make it clear how the product rule helps in separating the parts of the product and differentiating them.
Derivative
At the heart of calculus lies the concept of derivatives. A derivative tells us both the rate at which a function changes and the direction of that change. When we perform implicit differentiation, we are essentially finding the derivative, \(y'\), of a function that isn't in standard form, i.e., \(y = f(x)\).
  • Start differentiating all terms concerning \(x\), which in this exercise means recognizing \(y\) as a function of \(x\).
  • Use established techniques like the chain and product rules effectively to handle different types of terms like polynomials or products of functions.
  • Recollect that any constant has a derivative of zero, simplifying rightmost terms like \(5x + 3\) directly to \(5\).
  • Combine and rearrange the derivatives to find \(y'\).
Finally, we organize terms containing \(y'\) on one side of the equation and solve algebraically. This process gives us a clearer understanding of how solutions to equations change concerning their variables.

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Most popular questions from this chapter

Evaluate the integral. $$ \int \frac{1}{x-x \ln x} d x $$

One thousand trout, each one year old, are introduced into a large pond. The number still alive after \(t\) years is predicted to be \(N(t)=1000(0.9)^{t}\) (a) Approximate the death rate \(d N / d t\) at times \(t=1\) and \(t=5\). At what rate is the population decreasing when \(N=500 ?\) (b) The weight \(W(t)\) (in pounds) of an individual trout is expected to increase according to the formula \(W(t)=\) \(0.2+1.5 t .\) After approximately how many years is the total number of pounds of trout in the pond a maximum?

Solve the differential equation subject to the given conditions. $$ y^{\prime \prime}=3 e^{-x} ; \quad y=-1 \text { and } y^{\prime}=1 \text { if } x=0 $$

Evaluate the integral. $$ \int\left(1+e^{-3 x}\right)^{2} d x $$

If a savings bond is purchased for \(\$ 500\) with interest compounded continuously at \(7 \%\) per year, then after \(t\) years the bond will be worth \(500 e^{0.07 t}\) dollars. |al Approximately when will the bond be worth \(\$ 1000 ?\) |b| Approximately when will the value of the bond be growing at a rate of \(\$ 50\) per year?
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