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Find the derivatives of the functions in Exercises \(19-38\) $$ q=\sqrt{2 r-r^{2}} $$

Short Answer

Expert verified
The derivative is \( q' = \frac{1 - r}{\sqrt{2r - r^2}} \).

Step by step solution

01

Rewrite the Function

Before finding the derivative, rewrite the function to make it easier to differentiate. The function is \( q = \sqrt{2r - r^2} \). Rewrite this as \( q = (2r - r^2)^{1/2} \).
02

Apply the Chain Rule

To find the derivative, apply the chain rule. The chain rule states that if \( y = (g(x))^n \), then \( \frac{dy}{dx} = n(g(x))^{n-1} \cdot g'(x) \). Here, \( g(x) = 2r - r^2 \) and \( n = 1/2 \).
03

Differentiate the Outer Function

Apply the power rule to the outer function. According to the power rule, the derivative of \( (2r - r^2)^{1/2} \) is \( \frac{1}{2}(2r - r^2)^{-1/2} \).
04

Differentiate the Inner Function

Find the derivative of the inner function, \( 2r - r^2 \). The derivative is \( 2 - 2r \).
05

Combine Derivatives and Simplify

Combine the results from Steps 3 and 4 using the chain rule: \( q' = \frac{1}{2}(2r - r^2)^{-1/2} \cdot (2 - 2r) \). Simplify this expression to get: \( q' = \frac{2 - 2r}{2\sqrt{2r - r^2}} = \frac{1 - r}{\sqrt{2r - r^2}} \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental technique used in calculus to find the derivative of a composition of functions. Consider it as a method to tackle more complex differentiation problems that involve nested functions. It essentially breaks down the differentiation process into manageable parts.

To apply the chain rule, we look at the outer and inner functions separately. For example, when differentiating something like \(y = (g(x))^n\), we treat \(g(x)\) as the inner function and \(n\) as how we handle the outer power. The formula given by the chain rule is:
  • Outer Function Derivative: Differentiate the outer function but leave the inner function untouched.
  • Inner Function Derivative: Differentiate the inner function.
  • Combine: Multiply the results, following the chain rule formula \(\frac{dy}{dx} = n(g(x))^{n-1} \cdot g'(x)\).
In this exercise, our outer function was \( (2r - r^2)^{1/2}\), and after handling the outer component with the power rule, the chain rule allowed us to incorporate the inner function's derivative efficiently. This systematic approach ensures you handle even the most intricate functions smoothly.
Power Rule
The power rule is a straightforward and vital method used to find derivatives quickly and easily. It involves functions that are in the form \(x^n\), where \(n\) is any real number. The rule helps us take derivatives by reducing the power of the term by one and multiplying by the original power.

To apply the power rule, follow these simple steps:
  • Identify the power of the expression you are differentiating.
  • Bring down this power as a multiplier in front of the term.
  • Decrease the original power by one to find the new exponent.
For example, in the given exercise, the outer function \( (2r - r^2)^{1/2}\) was differentiated using the power rule. The power, \(1/2\), came down in front, and the new exponent became \(-1/2\). This transformation underpins the concept because it simplifies the process of finding derivatives and is often paired with the chain rule.

Utilizing these steps makes differentiation a systematic and less daunting task, enhancing our calculus toolkit.
Calculus Exercises
Calculus exercises are indispensable for anyone looking to master the subject of calculus. They provide a hands-on approach to understanding the abstract concepts taught in the classroom or found in textbooks. Through regular practice, students deepen their comprehension of differentiation and integration and improve their problem-solving skills.

For the exercise discussed, rewriting the function as \(q = (2r - r^2)^{1/2}\) is a common first step. This step allows us to see where we can use calculus rules like the chain and power rules. Exercises like these train students to:
  • Identify suitable techniques for various problems.
  • Apply rules systematically rather than guesswork.
  • Simplify expressions for clarity and accuracy.
Understanding how to transition from a complex function to its derivative equips students with the ability to tackle more challenging problems. Hence, regular practice with exercises is key to grasping calculus concepts effectively and preparing for exams or real-world applications.

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Most popular questions from this chapter

In Exercises 45 and \(46,\) find the slope of the curve at the given points. $$ \left(x^{2}+y^{2}\right)^{2}=(x-y)^{2} \quad \text { at } \quad(1,0) \text { and }(1,-1) $$

In Exercises \(87-94\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=\cos t, \quad y=\sqrt{3} \cos t, \quad t=2 \pi / 3 $$

In Exercises \(67-70\) , use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\) . Perform the following steps: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=\sqrt{x}-\sin x, \quad[0,2 \pi], \quad a=2 $$

Normals to a parabola Show that if it is possible to draw three normals from the point \((a, 0)\) to the parabola \(x=y^{2}\) shown here, then \(a\) must be greater than 1\(/ 2\) . One of the normals is the \(x\) -axis. For what value of \(a\) are the other two normals perpendicular?

a. Given that \(x^{4}+4 y^{2}=1\) , find \(d y / d x\) two ways: \((1)\) by solving for \(y\) and differentiating the resulting functions in the usual way and \((2)\) by implicit differentiation. Do you get the same result each way? b. Solve the equation \(x^{4}+4 y^{2}=1\) for \(y\) and graph the resulting functions together to produce a complete graph of the equation \(x^{4}+4 y^{2}=1 .\) Then add the graphs of the first derivatives of these functions to your display. Could you have predicted the general behavior of the derivative graphs from looking at the graph of \(x^{4}+4 y^{2}=1 ?\) Could you have predicted the general behavior of the graph of \(x^{4}+4 y^{2}=1\) by looking at the derivative graphs? Give reasons for your answers.

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