/*! 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 44 Find an equation of the tangent ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find an equation of the tangent line to the graph of the logarithmic function at the point \((1,0)\). $$ y=\ln x^{1 / 2} $$

Short Answer

Expert verified
The equation of the tangent line to the graph of the function at the point (1,0) is \(y = 1/2x - 1/2\).

Step by step solution

01

Differentiate the function

Start by differentiating the logarithmic function using the chain rule: \(dy/dx = (d/dx[\ln(x^{1/2})]) = (1/2x)(1/x) = 1/2x\).
02

Find the value of the derivative at the point (1,0)

Substitute \(x=1\) into the derivative to find the slope of the tangent line at the point (1,0) which gives: \(dy/dx = 1/2\).
03

Find the equation of the tangent line

Now use the point-slope form of a line equation: \(y- y1 = m(x - x1)\). Substitute the point (1,0) and the slope \(m = 1/2\) into this equation to find the equation of the tangent line, which gives: \(y - 0 = 1/2 (x - 1)\), and simplifies to \(y = 1/2x - 1/2\).

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Key Concepts

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

Differentiating Logarithmic Functions
Understanding how to differentiate logarithmic functions is crucial in calculus. These functions have the general form \( y = \ln(x^{{n}}) \) where \( n \) is a real number. To differentiate \( \ln(x^{{n}}) \) with respect to \( x \) we use the power rule in conjunction with the chain rule. A logarithmic function essentially measures the power to which a base, typically \( e \) (Euler's number), must be raised to obtain the number that is the argument of the logarithm.

In our example, the function \( y = \ln(x^{{1 / 2}}) \) can be differentiated by first rewriting it as \( y = \frac{1}{2}\ln(x) \) using logarithm properties. The derivative of \( \ln(x) \) with respect to \( x \) is \( 1/x \) and applying the constant multiple rule, we get \( dy/dx = \frac{1}{2}(1/x) \) or simply \( 1/(2x) \). This process demonstrates how combining various differentiation rules allows for the successful derivation of more complex functions.

Understanding and applying these concepts correctly is key to finding the rate of change of logarithmic functions at any given point, which is the essence of differentiation in calculus.
Chain Rule in Calculus
The chain rule is a fundamental tool in calculus for finding the derivative of a composition of functions. It states that if you have a function \( g \) which is composed of another function \( f \), represented as \( g(f(x)) \), then the derivative of \( g \) with respect to \( x \) is \( g'(f(x)) \times f'(x) \). This means you multiply the derivative of \( g \) with respect to \( f(x) \) by the derivative of \( f \) with respect to \( x \).

In the context of logarithmic functions, such as \( y = \ln(x^{1 / 2}) \), we identify \( g \) as the log function and \( f(x) \) as \( x^{1 / 2} \). Applying the chain rule, we determine the derivative of the outer function \( g \) with respect to \( f(x) \) and then multiply it by the derivative of \( f(x) \) with respect to \( x \). The chain rule simplifies the differentiation process, breaking down complex functions into manageable chunks that can be differentiated separately and then combined to find the overall derivative.
Point-Slope Form of a Line
The point-slope form is an equation used to describe the line that has a specific slope and passes through a given point. It is expressed as \( y - y_{{1}} = m(x - x_{{1}}) \) where \( m \) is the slope of the line and \( (x_{{1}}, y_{{1}}) \) is the point through which the line passes. This formula is particularly useful when writing an equation of a tangent line to a function at a given point.

When we talk about a tangent line to a curve at a certain point, we refer to a line that just touches the curve at that point and has the same slope as the curve at that point. The slope of the tangent line at any point on a function is simply the value of the derivative of that function at that point. Once we have the slope and we have a point on the line (in our case, \( (1,0) \) on the curve \( y = \ln(x^{1 / 2}) \) ), we can plug these into the point-slope equation to get the equation of our tangent line: \( y = 1/2x - 1/2 \). This equation now represents the tangent line at the point \( (1,0) \) on our original function, providing a visual and mathematical representation of the behavior of the function at that specific point.

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Most popular questions from this chapter

Find any relative extrema of the function. Use a graphing utility to confirm your result. $$ g(x)=x \operatorname{sech} x $$

Find the derivative of the function. $$ y=x \tanh ^{-1} x+\ln \sqrt{1-x^{2}} $$

Use the equation of the tractrix \(y=a \operatorname{sech}^{-1} \frac{x}{a}-\sqrt{a^{2}-x^{2}}, \quad a>0\). Find \(d y / d x\).

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

Find the integral. $$ \int \frac{\cosh \sqrt{x}}{\sqrt{x}} d x $$

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