Another method for proving lim x→0 cos x−1/x = 0 Use the half-angle formula sin²x = 1− cos 2x/2 to prove that lim x→0 cos x−1/x=0.
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.8.91
90–93. {Use of Tech} Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph.
x²(3y²−2y³) = 4
Verified step by step guidance1
Start by differentiating the given equation implicitly with respect to x. The equation is x²(3y²−2y³) = 4. Use the product rule for differentiation, which states that if you have a product of two functions u(x) and v(x), then the derivative is u'(x)v(x) + u(x)v'(x).
Differentiate the left side of the equation: For x²(3y²−2y³), let u = x² and v = 3y²−2y³. The derivative of u with respect to x is 2x, and the derivative of v with respect to x involves implicit differentiation: d/dx(3y²−2y³) = 6y(dy/dx) - 6y²(dy/dx).
Apply the product rule: The derivative of x²(3y²−2y³) is 2x(3y²−2y³) + x²(6y(dy/dx) - 6y²(dy/dx)). Set this equal to the derivative of the right side, which is 0, since the derivative of a constant (4) is 0.
To find points where the tangent line is horizontal, set dy/dx = 0 and solve the resulting equation for x and y. This will give you the conditions under which the slope of the tangent is zero.
To find points where the tangent line is vertical, identify where the expression for dy/dx becomes undefined. This typically occurs when the denominator of the expression for dy/dx is zero. Solve for x and y under these conditions, and verify that these points lie on the original curve by substituting back into the original equation.

Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
9mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. Instead of solving for one variable in terms of the other, we differentiate both sides of the equation with respect to the independent variable, applying the chain rule as necessary. This method is particularly useful for finding derivatives of curves defined by equations that cannot be easily rearranged.
Recommended video:
Finding The Implicit Derivative
Tangent Lines and Slopes
The slope of a tangent line to a curve at a given point represents the instantaneous rate of change of the curve at that point. A horizontal tangent line indicates a slope of zero, while a vertical tangent line suggests an undefined slope. To find these points, we set the derivative equal to zero for horizontal tangents and check where the derivative does not exist for vertical tangents.
Recommended video:
Slopes of Tangent Lines
Graphical Confirmation
Graphical confirmation involves plotting the curve and visually inspecting the points where the tangent lines are horizontal or vertical. This step is crucial as it helps verify the analytical results obtained through differentiation. By comparing the calculated points with the graph, one can ensure that the identified points indeed lie on the curve and accurately represent the behavior of the function.
Recommended video:
Determining Differentiability Graphically
Related Practice
Textbook Question
370
views
Textbook Question
Find the derivative of the following functions.
y = In(e^x + e^-x)
337
views
Textbook Question
The line tangent to the graph of f at x=5 is y = 1/10x-2. Find d/dx (4f(x)) |x+5
232
views
Textbook Question
The speed of sound (in m/s) in dry air is approximated the function v(T) = 331 + 0.6T, where T is the air temperature (in degrees Celsius). Evaluate v' (T) and interpret its meaning.
339
views
Textbook Question
Interpreting the derivative Find the derivative of each function at the given point and interpret the physical meaning of this quantity. Include units in your answer.
An object dropped from rest falls d(t)=16t² feet in t seconds. Find d′(4).
263
views
Textbook Question
Derivatives from graphs Use the figure to find the following derivatives. <IMAGE>
d/dx (xg(x)) | x=2
233
views
