Finding the Equation of a Tangent Line at a Given Point

In summary, the equation of the tangent line to the graph of the equation y = xe^(1/x^2) + ln(3 - 2x^2) at the point (1, e) is -e-4(x-1)+e.
  • #1
noboost4you
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I need to know how to find the equation of the tangent line to the graph of the equation y = xe^(1/x^2) + ln(3 - 2x^2) at the point (1, e). Admittedly, I have no idea at all what the question is asking for and I don't have the slighest clue on how to solve this question. Any help will be greatly appreciated. Thanks again
 
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  • #2
well if I remember correctly, you must compute the first derivative of the function y(x) in your point of interest [that is (1, e) in your case] and that will give you the slope of the tangent to the graph in that point.
After that you know the slope of the tangent and you also know one point of the tangent [that is also (1, e) in your case] and you can easily determine it's equation.
 
  • #3
Any straight line can be written in the form y= m(x-x0)+ y0 where m is the slope of the line and (x0,y0 is a point the line passes through.

The derivative, f', of a function f can be defined as the slope of the tangent line at each point.

In order to find the equation of the tangent line to y= f(x) at the point (x0,y0), find m= f'(x0) and write y= m(x-x0)+ y0.

In this problem f(x)= xe^(1/x^2) + ln(3 - 2x^2) so
f'= (1)e^(1/x^2)+ x(1/x^2)(-2/x^3)e^(1/x^2)+ (1/(3-2x^2))(-4x)
(Notice use of product rule and chain rule)

f(1)= e+ ln(1)= e so (1, e)is on the curve.
f'(1)= e- 2e- 4= -e-4

The equation of the tangent line to y= f(x) at (1,e) is
y= (-e-4)(x-1)+ e.
 
  • #4
Originally posted by HallsofIvy

In this problem f(x)= xe^(1/x^2) + ln(3 - 2x^2) so
f'= (1)e^(1/x^2)+ x(1/x^2)(-2/x^3)e^(1/x^2) + (1/(3-2x^2))(-4x)
(Notice use of product rule and chain rule)

The second part of that derivative is throwing me off. Wouldn't the derivative of xe^(1/x^2) be: (1)e^(1/x^2) + e^(1/x^2)(x)(-2/x^3) ?? I don't know how you incorporated (1/x^2) in that? Please shine some light on that problem. Thanks again
 
  • #5
Originally posted by noboost4you
The second part of that derivative is throwing me off. Wouldn't the derivative of xe^(1/x^2) be: (1)e^(1/x^2) + e^(1/x^2)(x)(-2/x^3) ?? I don't know how you incorporated (1/x^2) in that? Please shine some light on that problem. Thanks again

Well I guess that really wasn't that big of a deal because when f'(1) is plugged in, that extra (1/x^2) equals 1 anyhow. I just didn't see how that came into the derivative...
 

1. What is the equation of the tangent line?

The equation of the tangent line is a mathematical expression that represents the line that touches a curve at a specific point, called the point of tangency. It is used to approximate the behavior of a curve at a particular point.

2. How is the equation of the tangent line calculated?

The equation of the tangent line is calculated using the derivative of the function at the point of tangency. The derivative represents the slope of the curve at that point, and the equation of the tangent line can be written in the form y = mx + b, where m is the slope and b is the y-intercept.

3. What is the significance of the equation of the tangent line?

The equation of the tangent line is significant because it allows us to approximate the behavior of a curve at a specific point. It is also useful in finding the slope of a curve, which is an essential concept in calculus.

4. Can the equation of the tangent line be used to find the slope of a curve at any point?

Yes, the equation of the tangent line can be used to find the slope of a curve at any point. This is because the slope of a tangent line is equal to the derivative of the function at that point.

5. How is the equation of the tangent line used in real-life applications?

The equation of the tangent line is used in various real-life applications, such as physics, engineering, and economics. It is used to approximate the behavior of curves and to find the slope of a curve, which is crucial in understanding the rate of change of a system.

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