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

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SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
\[f(x)=\sqrt{x^2+16}\Rightarrow f(13)=\sqrt{185}\\ f'(x)=\frac{2x}{\sqrt{x^2+16}}\Rightarrow f'(13)=\frac{26}{\sqrt{185}}\] The tangent line is given by the pointslope formula using the point (13, √185) and the slope 26/√185: \[y\sqrt{185}=\frac{26}{\sqrt{185}}(x13)\]
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
so thats the equation? @SithAndGiggles
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
@SithsAndGiggles
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
the derivative is wrong. The nominator should be just x
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
okay u mean in 2x?
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
x/sqrt(x^2+16)
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
so it wouldn't be 26
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
rest looks right, just change 26 for 13 and that's it
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Thanks for catching the mistake
 one year ago
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