anonymous
  • anonymous
Find an equation of the tangent line to y = f(x) at x = a. f(x) = sqrt(x^2 + 16), x = 13
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[f(x)=\sqrt{x^2+16}\Rightarrow f(13)=\sqrt{185}\\ f'(x)=\frac{2x}{\sqrt{x^2+16}}\Rightarrow f'(13)=\frac{26}{\sqrt{185}}\] The tangent line is given by the point-slope formula using the point (13, √185) and the slope 26/√185: \[y-\sqrt{185}=\frac{26}{\sqrt{185}}(x-13)\]
anonymous
  • anonymous
so thats the equation? @SithAndGiggles
anonymous
  • anonymous
@SithsAndGiggles

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anonymous
  • anonymous
@myko
anonymous
  • anonymous
the derivative is wrong. The nominator should be just x
anonymous
  • anonymous
okay u mean in 2x?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok
anonymous
  • anonymous
x/sqrt(x^2+16)
anonymous
  • anonymous
okay
anonymous
  • anonymous
so it wouldn't be 26
anonymous
  • anonymous
rest looks right, just change 26 for 13 and that's it
anonymous
  • anonymous
okay thx
anonymous
  • anonymous
yw
anonymous
  • anonymous
Thanks for catching the mistake

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