/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 In your own words, state the gui... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In your own words, state the guidelines for implicit differentiation.

Short Answer

Expert verified
The guidelines for implicit differentiation involve differentiating all terms as normal but applying the chain rule for terms not with respect to the desired variable, then rearranging to isolate the desired derivative, and finally simplifying the result.

Step by step solution

01

Differentiating Both Sides

Start by differentiating every term on both sides of the equation with respect to the desired variable. The process is identical to explicit differentiation, however, remember that if a term includes another variable, it is also being implicitly varied by the variable you are differentiating with respect to.
02

Applying the Chain Rule

Whenever you differentiate a function that is not initially in terms of the variable you are differentiating with respect to, you must apply the chain rule. This means after normal differentiation, you must add a term that is the derivative of the function's current variable with respect to the variable you are differentiating with respect to. Always remember to apply the chain rule whenever necessary.
03

Rearranging to Solve for Desired Derivative

After differentiating, if you want to find the derivative of one variable with respect to another, you may need to rearrange to isolate the desired derivative. Use your algebra skills to move terms around so that all items involving the derivative of interest are on one side and the rest are on the other.
04

Simplifying the Result

Lastly simplify the implicitley derived derivative. This can involve cancelling out similar terms, or simplifying fractions or any algebraic simplification that makes the derived function, easier to understand and use.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chain Rule
When you're differentiating functions, sometimes you come across expressions that aren't explicitly in terms of your variable of interest. This is where the chain rule becomes an invaluable tool. Imagine you're peeling layers off an onion. Similarly, in implicit differentiation, you peel "layers" off by taking the derivative of each outer function in succession.
For example, if you are differentiating with respect to \(x\), and you encounter \(y\), which is a function of \(x\), you apply the chain rule. First, differentiate \(y\) as is, and then multiply by the derivative of \(y\) with respect to \(x\), denoted as \(\frac{dy}{dx}\).
This adjustment accounts for the changes in one variable due to changes in another. By always applying the chain rule when needed, you ensure that your differentiation is accurate and comprehensive.
  • Key point: Use the chain rule when differentiating a composite function.
  • Remember: After differentiating, multiply by the derivative of the inside variable with respect to the one you're differentiating.
Differentiation
Differentiation is the process used to find the rate at which a function is changing at any given point. In calculus, this is expressed as a derivative.
When applying implicit differentiation, the principles are the same as explicit differentiation, but with more complexity due to hidden variables inside another function. Start by taking the derivative of each term in an equation. If these terms involve a variable you are not directly differentiating with respect to, the chain rule will be required.
The goal of this process is to identify how these hidden variables are changing relative to each other. It provides a powerful way to handle relationships that are not explicitly stated, allowing the extraction of derivatives even when the function isn't neatly solved for one variable in terms of another.
  • Important: Consistently apply the principles of differentiation to each term.
  • Mistakes to avoid: Forgetting to apply the chain rule when necessary.
Algebra
Algebra comes into play after differentiating an implicit function. Once derivatives are found, algebraic manipulation is key to rearranging terms so the derived expression can be isolated and simplified. This often involves moving terms across the equation, factoring, or simplifying fractions.
For example, if you've differentiated an equation and your derivative, such as \(\frac{dy}{dx}\), is tangled within multiple terms, you'll use algebra to get it alone on one side. This makes the function easier to understand and use. Simplification ensures that the expression is in its simplest form, which can turn a complex result into a more manageable one.
Always remember:
  • Use algebraic principles like combining like terms or factoring.
  • Constantly check your work for opportunities to simplify.
By applying these algebraic techniques, you'll make the daunting task of implicit differentiation much more approachable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find equations for the tangent line and normal line to the circle at the given points. (The normal line at a point is perpendicular to the tangent line at the point.) Use a graphing utility to graph the equation, tangent line, and normal line. $$ \begin{aligned} &x^{2}+y^{2}=25 \\ &(4,3),(-3,4) \end{aligned} $$

Find equations of the tangent lines to the graph of \(f(x)=\frac{x+1}{x-1}\) that are parallel to the line \(2 y+x=6\). Then graph the function and the tangent lines.

The stopping distance of an automobile, on dry, level pavement, traveling at a speed \(v\) (kilometers per hour) is the distance \(R\) (meters) the car travels during the reaction time of the driver plus the distance \(B\) (meters) the car travels after the brakes are applied (see figure). The table shows the results of an experiment. $$\begin{array}{|l|c|c|c|c|c|}\hline \text { Speed, } \boldsymbol{v} & 20 & 40 & 60 & 80 & 100 \\\\\hline \begin{array}{l}\text { Reaction Time } \\\\\text { Distance, } \boldsymbol{R}\end{array} & 8.3 & 16.7 & 25.0 & 33.3 & 41.7 \\\\\hline \begin{array}{l}\text { Braking Time } \\\\\text { Distance, } \boldsymbol{B}\end{array} & 2.3 & 9.0 & 20.2 & 35.8 & 55.9 \\\\\hline\end{array}$$ (a) Use the regression capabilities of a graphing utility to find a linear model for reaction time distance. (b) Use the regression capabilities of a graphing utility to find a quadratic model for braking distance. (c) Determine the polynomial giving the total stopping distance \(T\). (d) Use a graphing utility to graph the functions \(R, B\), and \(T\) in the same viewing window. (e) Find the derivative of \(T\) and the rates of change of the total stopping distance for \(v=40, v=80\), and \(v=100\) (f) Use the results of this exercise to draw conclusions about the total stopping distance as speed increases.

Let \(A\) be the area of a circle of radius \(r\) that is changing with respect to time. If \(d r / d t\) is constant, is \(d A / d t\) constant? Explain.

Determine the point(s) in the interval \((0,2 \pi)\) at which the graph of \(f(x)=2 \cos x+\sin 2 x\) has a horizontal tangent.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.