Find an equation for the tangent line to the curve at point (2,1)

A

alexmay

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y= 5/x^2 +1

I need help approaching this problem, and how to go about it step by step.
 
It appears what you meant is:

[MATH]y=\frac{5}{x^2+1}[/MATH]
What is [MATH]\d{y}{x}[/MATH]?
 
Please draw a diagram. You can draw any curve you want for y, except a straight line.

Find the point on this curve whose x coordinate is 1. Now draw the tangent line.

To get the equation of a line you can do this if you have a point and slope.

The key here is to know what information the derivative ,y', gives you. If you plug in for example 7 for x in the y' (NOT y) formula you will get the slope of the tangent line to the y curve where the x coordinations is 7
 
alexmay said:
I'm not sure
Click to expand...

I'd be tempted to write:

[MATH]y=5(x^2+1)^{-1}[/MATH]
And then apply the power/chain rules, rather than use the quotient rule on the original form. But either will work. Can you post some work?
 
MarkFL said:
I'd be tempted to write:

[MATH]y=5(x^2+1)^{-1}[/MATH]
And then apply the power/chain rules, rather than use the quotient rule on the original form. But either will work. Can you post some work?
Click to expand...
I agree with your approach but shouldn't we wait for the OP to confirm that we even know the correct function?
 
Jomo said:
I agree with your approach but shouldn't we wait for the OP to confirm that we even know the correct function?
Click to expand...

I assumed the form I gave rather than what was presented since the given point exists on the form I gave and not on the other. :)
 
MarkFL said:
I assumed the form I gave rather than what was presented since the given point exists on the form I gave and not on the other. :)
Click to expand...
Oops I missed the given point. Why do students put some information in the heading? I honestly thought the equation was \(\displaystyle \frac{5}{x^2}+1\). Thanks!
 
You need to know that \(\displaystyle \dfrac{d}{dx} (f(x))^n = n(f(x))^{n-1}*f'(x)\)

What is your f(x) and n? How do you handle that 5 in the front?
 
With:

[MATH]y=5(x^2+1)^{-1}[/MATH]
We find:

[MATH]\d{y}{x}=-5(x^2+1)^{-2}(2x)=-\frac{10x}{(x^2+1)^{2}}[/MATH]
Hence

[MATH]\left.\d{y}{x}\right|_{x=2}=-\frac{10(2)}{(2^2+1)^{2}}=-\frac{4}{5}[/MATH]
And so the tangent line is:

[MATH]y=-\frac{4}{5}(x-2)+1=-\frac{4}{5}x+\frac{13}{5}\quad\checkmark[/MATH]
 
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