/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 Find the tangent to \(y=((x-1) /... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the tangent to \(y=((x-1) /(x+1))^{2}\) at \(x=0\)

Short Answer

Expert verified
The tangent line at \( x = 0 \) is \( y = -4x + 1 \).

Step by step solution

01

Find the derivative

To find the equation of the tangent line, we first need to find the derivative of the function. Let \( y = \left( \frac{x-1}{x+1} \right)^2 \). To find \( \frac{dy}{dx} \), use the chain rule combined with the quotient rule. \( u = \frac{x-1}{x+1} \), hence \( y = u^2 \). \( \frac{dy}{du} = 2u \) and \( \frac{du}{dx} = \frac{(x+1)(1)-(x-1)(1)}{(x+1)^2} = \frac{2}{(x+1)^2} \). Thus, \( \frac{dy}{dx} = 2 \cdot \frac{x-1}{x+1} \cdot \frac{2}{(x+1)^2} = \frac{4(x-1)}{(x+1)^3} \).
02

Evaluate the derivative at the given point

Now, we evaluate the derivative at \( x = 0 \) to find the slope of the tangent line. Substituting \( x = 0 \) into \( \frac{dy}{dx} = \frac{4(x-1)}{(x+1)^3} \) yields \( \frac{dy}{dx}\bigg|_{x=0} = \frac{4(0-1)}{(0+1)^3} = \frac{-4}{1} = -4 \).
03

Find the y-coordinate of the point of tangency

Substitute \( x = 0 \) into the original function to find the y-coordinate of the point where the tangent touches the curve: \( y = \left( \frac{0-1}{0+1} \right)^2 = 1 \). Thus, the point of tangency is \( (0, 1) \).
04

Write the equation of the tangent line

With the slope \( m = -4 \) and the point of tangency \( (0, 1) \), we use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \). Plugging in the values, \( y - 1 = -4(x - 0) \), which simplifies to \( y = -4x + 1 \).

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Key Concepts

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

Understanding Derivatives
Derivatives are fundamental in calculus, providing a powerful tool to understand how functions change. In the context of curves, the derivative represents the slope of the tangent line, which touches the curve at a single point. This slope shows how steep the curve is at that particular moment. A derivative is often denoted by \( \frac{dy}{dx} \), which signifies the rate of change of \( y \) with respect to \( x \).
Imagine you're hiking up a hill. At each step, the steepness can vary - a derivative tells you exactly how steep your path is at every point.
Finding the derivative of complex functions may pose challenges, but rules such as the chain and quotient rules simplify the process, allowing us to efficiently calculate the derivative.
The Point-Slope Form Unveiled
Once you have the slope of a tangent line, the next step is expressing this line in equation form. The point-slope form is a straightforward method for this purpose. It’s written as:
  • \( y - y_1 = m(x - x_1) \)
Here, \( m \) is the slope, and \( (x_1, y_1) \) is a specific point on the line.
Imagine point-slope form as a simple recipe. You mix the slope, like a magical ingredient, with a point, like a key flavor, to get the complete equation.
This formula is particularly useful when dealing with tangent lines, as these lines often intersect the curve at a known point with a precise slope. With the steps above, we found the slope at \( x = 0 \), and using the point \((0,1)\), we crafted the simple equation of the tangent line.
Mastering the Chain Rule
The chain rule is a very handy tool when dealing with composite functions, or functions within another function. It allows us to differentiate complex expressions by breaking them down into manageable parts.
For instance, suppose you have \( y = (u(x))^2 \), where \( u(x) \) is itself a function. The chain rule helps by letting you find the derivative of \( y \) with respect to \( x \) through an intermediate function \( u \).
The formula is:
  • \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)
In our example, we applied the chain rule in combination with the quotient rule as we managed a quotient raised to a power. This approach allows us to handle the inner function first, then address the outer function, as demonstrated in computing the derivative of \( y = \left( \frac{x-1}{x+1} \right)^2 \). Thus, the chain rule provides a structured pathway for tackling derivatives of layered functions.

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Most popular questions from this chapter

Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=\) \(m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

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A particle moves along the \(x\) -axis with velocity \(d x / d t=f(x) .\) Show that the particle's acceleration is \(f(x) f^{\prime}(x)\)

The formula \(V=k r^{4},\) discovered by the physiologist Jean Poiseuille (1797-1869), allows us to predict how much the radius of a partially clogged artery has to be expanded in order to restore normal blood flow. The formula says that the volume \(V\) of blood flowing through the artery in a unit of time at a fixed pressure is a constant \(k\) times the radius of the artery to the fourth power. How will a \(10 \%\) increase in \(r\) affect \(V ?\)

Suppose that the functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=0\) and \(x=1\) $$\begin{array}{ccccc} \hline x & f(x) & g(x) & f^{\prime}(x) & g^{\prime}(x) \\ \hline 0 & 1 & 1 & 5 & 1 / 3 \\ 1 & 3 & -4 & -1 / 3 & -8 / 3 \\ \hline \end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x\) a. \(5 f(x)-g(x), \quad x=1\) b. \(f(x) g^{3}(x), \quad x=0\) c. \(\frac{f(x)}{g(x)+1}, \quad x=1\) d. \(f(g(x)), \quad x=0\) e. \(g(f(x)), \quad x=0\) f. \(\left(x^{11}+f(x)\right)^{-2}, \quad x=1\) g. \(f(x+g(x)), \quad x=0\)

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