onegirl

Find an equation of the tangent line to y = f(x) at x = a. f(x) = sqrt(x^2 + 16), x = 13

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SithsAndGiggles

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\[f(x)=\sqrt{x^2+16}\Rightarrow f(13)=\sqrt{185}\\
f'(x)=\frac{2x}{\sqrt{x^2+16}}\Rightarrow f'(13)=\frac{26}{\sqrt{185}}\]
The tangent line is given by the point-slope formula using the point (13, √185) and the slope 26/√185:
\[y-\sqrt{185}=\frac{26}{\sqrt{185}}(x-13)\]

onegirl

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so thats the equation? @SithAndGiggles

onegirl

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@SithsAndGiggles

onegirl

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@myko

myko

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the derivative is wrong. The nominator should be just x

onegirl

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okay u mean in 2x?

myko

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yes

onegirl

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ok

myko

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x/sqrt(x^2+16)

onegirl

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okay

onegirl

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so it wouldn't be 26

myko

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rest looks right, just change 26 for 13 and that's it

onegirl

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okay thx

myko

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yw

SithsAndGiggles

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Thanks for catching the mistake