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Chapter 3: Problem 36

a. Find an equation of the line tangent to the given curve at a. b. Use a graphing utility to graph the curve and the tangent line on the same set of axes. $$y=\frac{e^{x}}{x} ; a=1$$

Short Answer

Expert verified
Answer: The equation of the tangent line at $$x=1$$ is $$y=e$$. This is obtained by differentiating the given function, finding the slope of the tangent line at the given point, and using the point-slope form of a linear equation to find the equation of the tangent line.

Step by step solution

01

Differentiate the function

Differentiate the function $$y=\frac{e^x}{x}$$ with respect to $$x$$ using the quotient rule, which states that if $$u(x)$$ and $$v(x)$$ are functions of $$x$$, then: $$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vd(u)-u(dv)}{v^2}$$ Here, $$u(x)=e^x$$ and $$v(x)=x$$, thus: $$u'(x)=e^x$$ and $$v'(x)=1$$. Applying the quotient rule gives us: $$\frac{dy}{dx}=\frac{x(e^x)-e^x(1)}{x^2}$$
02

Find the slope of the tangent line

To find the slope of the tangent line, plug in the value $$x=1$$ into the derivative: $$\frac{d}{dy}\left(\frac{e^1}{1}\right)=\frac{1(e^1)-e^1(1)}{1^2}=e^1-e^1=0$$
03

Find the y-coordinate of the point

To find the $$y$$-coordinate of the point where the tangent line meets the curve, plug in the value $$x=1$$ into the original function: $$y(1)=\frac{e^1}{1}=e$$
04

Find the equation of the tangent line

Now, we have the point $$(1,e)$$ and the slope $$0$$, we can use the point-slope form of a linear equation to find the equation of the tangent line, which is: $$y-y_1=m(x-x_1)$$ Where $$(x_1,y_1)$$ is the point on the tangent line, and $$m$$ is the slope. Plugging in the values we have: $$y-e=0(x-1)$$ Simplifying this equation gives us the equation of the tangent line: $$y=e$$ The equation of the tangent line at $$x=1$$ is $$y=e$$. To graph the given curve and tangent line, use a graphing utility.

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