Chapter 3: Problem 11
Find and compare the values of \(d y\) and \(\Delta y\) for each function at the given values of \(x\) and \(d x=\Delta x\). \(y=\frac{3 x+5}{x^{2}+1}\) at \(x=1\) and \(d x=\Delta x=0.2\)
Short Answer
Expert verified
\(dy = -0.1\), \(\Delta y \approx -0.4754\).
Step by step solution
01
Differentiate the Function
To find \(dy\), we need to calculate the derivative of \(y = \frac{3x+5}{x^2+1}\). Using the quotient rule, the derivative \(y'\) is \[ y' = \frac{(x^2+1)(3) - (3x+5)(2x)}{(x^2+1)^2} \].
02
Simplify the Derivative
Simplify the expression found from the quotient rule: \[ y' = \frac{3x^2 + 3 - 6x^2 - 10x}{(x^2+1)^2} = \frac{-3x^2 - 10x + 3}{(x^2+1)^2} \].
03
Calculate dy
Plug in \(x = 1\) and \(dx = 0.2\) into the derivative to find \(dy\): \[ dy = y'(1) \cdot dx = \frac{-3(1)^2 - 10(1) + 3}{(1^2+1)^2} \cdot 0.2 = \frac{-10}{4} \cdot 0.2 = -0.5 \cdot 0.2 = -0.1 \].
04
Calculate Delta y
To find \(\Delta y\), evaluate the original function at \(x = 1\) and \(x = 1 + 0.2 = 1.2\). First, evaluate \(y(1)\): \[ y(1) = \frac{3(1) + 5}{1^2 + 1} = \frac{8}{2} = 4 \].
05
Calculate y at x = 1.2
Now evaluate \(y(1.2)\): \[ y(1.2) = \frac{3(1.2) + 5}{(1.2)^2 + 1} = \frac{3.6 + 5}{1.44 + 1} = \frac{8.6}{2.44} \approx 3.5246 \].
06
Find Delta y
Calculate \(\Delta y = y(1.2) - y(1) = 3.5246 - 4 = -0.4754 \).
07
Compare dy and Delta y
Finally, compare \(dy\) and \(\Delta y\): \(dy = -0.1\) and \(\Delta y \approx -0.4754\). Though \(dy\) and \(\Delta y\) are both negative, their magnitudes are significantly different.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule is a method for finding the derivative of a function that is the division of two differentiable functions. When dealing with the function of the form \( y = \frac{u(x)}{v(x)} \), the derivative \( y' \) is given by the formula: \[ y' = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \] where \( u(x) \) and \( v(x) \) are both functions of \( x \), and \( u'(x) \) and \( v'(x) \) are their respective derivatives.
- The numerator of this formula is the difference between two terms: the derivative of the numerator function, \( u(x) \), multiplied by the denominator, \( v(x) \), minus the numerator times the derivative of the denominator.
- Meanwhile, the denominator is simply the square of the denominator function \( v(x) \).
Derivative
Derivatives are a fundamental concept in differential calculus that measure how a function changes as its input changes. The derivative of a function \( f(x) \), denoted \( f'(x) \) or \( \frac{df}{dx} \), represents the slope of the tangent line to the curve at a given point \( x \).
- The process of finding a derivative is known as differentiation.
- Derivatives allow us to determine the rate at which one quantity changes with respect to another.
Differentials
Differentials provide an approximation of how much a function changes based on a small change in the input. When we have a function \( y = f(x) \) and its derivative \( y' \), the differential \( dy \) is defined as \[ dy = y'(x) \, dx \] where \( dx \) is a small change in \( x \).
- To compute \( dy \), you first find the derivative \( y'(x) \) of the function.
- Then multiply it by the small change \( dx \).
Approximation
Approximation in calculus often refers to estimating values of a function using its derivative. Differentials provide a linear approximation to a function by using the derivative:
- This approach assumes that changes in \( y \) over small \( x \) increments can be approximated linearly.
- Such techniques are very useful in applied mathematics and sciences where exact values might be difficult or impossible to obtain quickly.