Implicit Differentiation Tangent Line
Implicit differentiation is a technique used to find the derivative of an implicitly defined function. When finding the equation of a tangent line to a curve defined implicitly, implicit differentiation is often used to find the slope of the tangent line. This involves differentiating both sides of the implicit equation with respect to the independent variable and then solving for the derivative.
6 Key excerpts on "Implicit Differentiation Tangent Line"
eBook - ePub- Adil H. Mouhammed(Author)
- 2020(Publication Date)
- Routledge(Publisher)
...But if the function is written as g(x,y) = 0, then the function is said to be an implicit function. And the differentiation of such a function is called implicit differentiation. Here, we treat dy/dx as unknown and differentiate the function with respect to x. Having done so, we solve for dy/dx. Example 1: Differentiate xy = 9 implicitly. To differentiate this function implicitly with respect to x, we do the following. Rewrite the function as xy − 9 = 0, and differentiate with respect to x: the derivative of x with respect to x (which is one) times the second variable y, then the derivative of y with respect to x (which is dy/dx) times the first variable x. This gives d y / d x = y + [ dy / dx ] x = 0. Solving for dy/dx, we obtain dy/ dx = − y/x. Example 2: Differentiate x 2 + y 2 − 20 = 0 implicitly with respect to x. Solution: 2x + 2y dy/dx = 0. Thus, dy/dx = − x/y. Example 3: Differentiate xy − y + 4x = 0 implicitly with respect to x. Solution: y + (dy/dx) x − (dy/dx) + 4 = 0. Solving for dy/dx yields d x / d y = − (y + 4) / (x − 1). Example 4 Consumer Equilibrium. A consumer can achieve equilibrium if the indifference curve is tangent to the budget line, that is, if the slope of the indifference curve is equal to the slope of the budget line. This tangency point gives the number of units of commodities x and y that our consumer should buy to maximize his or her utility subject to the amount of income (budget). Suppose the indifference curve equation is X 0.5 Y 0.5 = 15. Let the budget equation be 2x + 3y = M, where x and y are two commodities, and the prices of x and y are $2 and $3, respectively; and x is the total income the consumer has. Given this information, we want to find how many units of x and y our consumer should buy. Solution: Obtain dy/dx, the marginal rate of substitution of x for y, from the indifference equation as well as from the budget equation, then equate the results and solve for x and y...
eBook - ePub- Gregory Hill, Mel Friedman(Authors)
- 2013(Publication Date)
- Research & Education Association(Publisher)
...The second subtlety is remembering to use the product or quotient rule whenever a product or quotient of x and y is encountered. For instance, in the implicit equation x 2 y 2 + 2 x = y 3, the term x 2 y 2 is a product of two functions of x, x 2 and y 2, and would require the use of the product rule. One way to gain confidence in the process is to examine an example that can be differentiated in either its explicit or its implicit form and see that the results are indeed the same. EXAMPLE 4.20 Given sin(x) · y = 4, find SOLUTION Isolating y leads to or y = 4 csc(x), so Now we use implicit differentiation with the product rule. Solving for yields Since y = 4 csc(x), substituting for y produces Now let’s examine the equation given in the previous paragraph to see how the product rule works in implicit differentiation. EXAMPLE 4.21 Given x 2 y 2 + 2 x = y 3, find in terms of x and y. SOLUTION Notice the use of the product rule on the left side of the equation and the power rule on the right side of the equation. The term is produced by differentiating the base of each power. of y. it is the chain rule factor. Notice that the derivative value now relies on both the x and y values at a given point. Sometimes when finding the value of the derivative at a point, one may have to actually use the entire ordered pair instead of just the abscissa. EXAMPLE 4.22 Find the equation of the line tangent to in the first quadrant at y = 1 SOLUTION First solve for x when y = 1 Keep x = 1 because the problem requires a first quadrant point. The point is (1, 1). Find to get the slope. The term requires using the quotient rule. Multiply each term by y 2. The point-slope form of the equation is The implicit curve and the line tangent at (1, 1) are shown in Figure 4.2. Figure 4.2 4.9 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS Implicit differentiation enables the exploration of the derivatives of an additional set of functions, inverse trigonometric functions...
- J. Rosebush, Stu Schwartz(Authors)
- 2016(Publication Date)
- Research & Education Association(Publisher)
...PART III DERIVATIVES Chapter 7 Derivatives I. DERIVATIVES A. Meaning of Derivative The derivative of a function is its slope. A linear function has a constant derivative since its slope is the same at every point. The derivative of a function at a point is the slope of its tangent line at that point. Non-linear functions have changing derivatives since their slopes (slope of their tangent line at each point) change from point to point. 1. Local linearity or linearization—when asked to find the linearization of a function at a given x -value or when asked to find an approximation to the value of a function at a given x -value using the tangent line, this means finding the equation of the tangent line at a “nice” x -value in the vicinity of the given x -value, substituting the given x -value into it and solving for y. i. For example, approximate using the equation of a tangent line to. We’ll find the equation of the tangent line to at x = 4 (this is the ‘nice’ x -value mentioned earlier). What makes it nice is that it is close to 4.02 and that. Since, so,. Also, f (4) = 2. Substituting these values into the equation of the tangent line, so the equation of the tangent line is. Substituting x = 4.02, y = 2.005. A more accurate answer (using the calculator) is. The linear approximation, 2.005, is very close to this answer. This works so well because the graph and its tangent line are very close at the point of tangency, thus making their y -values very close as well. If you use the tangent line to a function at x = 4 to approximate the function’s value at x = 9, you will get a very poor estimate because at x = 9, the tangent line’s y -values are no longer close to the function’s y -values. ii. The slope of the secant on (a, b), is often used to approximate the value of the slope at a point inside (a, b). For instance, given the table of values of f (x) below, and given that f (x) is continuous and differentiable, approximate f ′(3)...
eBook - ePub- Adil H. Mouhammed(Author)
- 2015(Publication Date)
- Routledge(Publisher)
...To differentiate this function implicitly with respect to x, we do the following. Rewrite the function as xy − 9 = 0, and differentiate with respect to x: the derivative of x with respect to x (which is one) times the second variable y, then the derivative of y with respect to x (which is dy/dx) times the first variable x. This gives dy/dx = y + [dy/dx] x = 0. Solving for dy/dx, we obtain dy/dx = - y/x. Example 2: Differentiate x 2 + y 2 − 20 = 0 implicitly with respect to x. Solution: 2x + 2y dy/dx = 0. Thus, dy/dx = − x/y. Example 3: Differentiate xy − y + 4x = 0 implicitly with respect to x. Solution: y + (dy/dx) x − (dy/dx) + 4 = 0. Solving for dy/dx yields dy/dx = - (y + 4) / (x - 1). Example 4: Consumer Equilibrium. A consumer can achieve equilibrium if the indifference curve is tangent to the budget line, that is, if the slope of the indifference curve is equal to the slope of the budget line. This tangency point gives the number of units of commodities x and y that our consumer should buy to maximize his or her utility subject to the amount of income (budget). Suppose the indifference curve equation is X 05 Y 05 = 15. Let the budget equation be 2x + 3y = M, where x and y are two commodities, and the prices of x and y are $2 and $3, respectively; and M is the total income the consumer has. Given this information, we want to find how many units of x and y our consumer should buy. Solution: Obtain dy/dx, the marginal rate of substitution of x for y, from the indifference equation as well as from the budget equation, then equate the results and solve for x and y...
eBook - ePub- Mike Aitken, Bill Broadhurst, Stephen Hladky(Authors)
- 2009(Publication Date)
- Garland Science(Publisher)
...As in these examples, we often start out with information about the derivative of a function without prior knowledge of the equation from which the derivative is derived. The name given to the process of reconstructing an original function from its derivative is integration. In other words, integration is the operation that must be performed to undo the effects of differentiation. We can start to unpack this concept by differentiating the function y = x 2, to achieve the familiar result that d y d x = 2 x. Integration is the reverse of this process, so alongside stating that ‘the derivative of x 2 is 2 x ’, it looks like we should also be able to say ‘the integral of 2 x is x 2 ’. Unfortunately, this picture is a little too simplistic. The true situation is more complex because an infinite number of functions can in fact be differentiated to get a result of 2 x. For example, we would get the same result if instead of y = x 2 we had chosen y = x 2 + 3 or even y = x 2 – 14. In each case, the derivatives of the second terms on the right-hand sides of the equations are zero, because they are all constants. Writing out these derivatives explicitly can emphasize the ambiguity of attempting to undo the effects of differentiation: d y d x (x 2) = 2 x ; d d x (x 2 + 3) = 2 x ; d y d x (x 2 − 14) = 2 x. In any attempt to reverse the process of differentiation, we have no idea what the value of the constant term in the original function may have been. Because of this, it is important to include an unknown constant in our answer for the integral of a function. The unknown constant, often written as c, is called the ‘constant of integration’. This means that the result for the integral of 2 x stated above was only partly correct...
- F. Xavier Malcata(Author)
- 2020(Publication Date)
- Wiley(Publisher)
...10 Differentials, Derivatives, and Partial Derivatives The concept of differential entails a small (tendentially negligible) variation in a variable x, denoted as dx, or a function f { x }, denoted as df { x }; the associated derivative of f { x } with regard to x is nothing but the ratio of said differentials, i.e. df / dx – usually known as Leibnitz’s formulation. In the case of a bivariate function, say, f { x,y }, differentials can be defined for both independent variables, i.e. dx and dy – so partial derivatives will similarly arise, i.e. ∂f / ∂x and ∂f / ∂y ; operator ∂ is equivalent to operator d, except that its use is exclusive to multivariate functions – in that it stresses existence of more than one independent variable. 10.1 Differential In calculus, the differential represents the principal part of the change of a function y =. f { x } – and its definition reads (10.1) where df / dx denotes the derivative of f { x } with regard to x ; it is normally finite, rather than infinitesimal or infinite – yet the precise meaning of variables dx and df depends on the context of application, and the required level of mathematical accuracy. The concept of differential was indeed introduced via an intuitive (or heuristic) definition by Gottfried W. Leibnitz, a German polymath and philosopher of the eighteenth century; its use was widely criticized until Cauchy defined it based on the derivative – which took the central role thereafter, and left dy free for given dx and df / dx as per Eq. (10.1). A graphical representation of differential is conveyed by Fig. 10.1, and the usefulness of differentials to approximate a function becomes clear from inspection thereof; after viewing dy as a small variation in the vertical direction, viz. (10.2) one may retrieve Eq...
