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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.49
Chapter 3, Problem 3.49

In Exercises 43–50, find by implicit differentiation.


y² = x .
x + 1

Verified step by step guidance
1
Start by differentiating both sides of the equation with respect to x. The equation is y² = x(x + 1).
Apply the product rule to the right side of the equation. The product rule 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). Here, u(x) = x and v(x) = (x + 1).
Differentiate the left side of the equation with respect to x. Since y is a function of x, use the chain rule: the derivative of y² with respect to x is 2y(dy/dx).
After differentiating both sides, you will have an equation involving dy/dx. Solve this equation for dy/dx to find the derivative of y with respect to x.
Simplify the expression for dy/dx, if possible, to get the final form of the derivative.

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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 equations that are difficult or impossible to rearrange.
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Finding The Implicit Derivative

Chain Rule

The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. It states that if a function y is composed of another function u, which in turn is a function of x, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x. This is essential in implicit differentiation when dealing with terms involving y.
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Intro to the Chain Rule

Derivative of a Function

The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In the context of implicit differentiation, understanding how to compute derivatives is crucial for finding the slope of the tangent line to the curve defined by the implicit equation.
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Derivatives of Other Trig Functions
Related Practice
Textbook Question

For Exercises 55 and 56, evaluate each limit by first converting each to a derivative at a particular x-value.


lim (x → 1) (x⁵⁰ − 1) / (x − 1)

Textbook Question

Derivatives in Differential Form


In Exercises 17–28, find dy.


y = sin(5√x)

Textbook Question

Second Derivatives


In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.


2√y = x – y

Textbook Question

In Exercises 41–58, find dy/dt.


y = (1 + cos(2t))⁻⁴

Textbook Question

Find the first and second derivatives of the functions in Exercises 33–38.


s = (t² + 5t − 1) / t²

Textbook Question

Is there a value of c that will make


f(x) = { (sin²(3x)) / x², x ≠ 0

c, x = 0


continuous at x = 0? Give reasons for your answer.