Question

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

Answer 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 point-slope formula using the point (13, √185) and the slope 26/√185:
\[y-\sqrt{185}=\frac{26}{\sqrt{185}}(x-13)\]

Answer 2

so thats the equation? @SithAndGiggles

Answer 3

@SithsAndGiggles

Answer 4

@myko

Answer 5

the derivative is wrong. The nominator should be just x

Answer 6

okay u mean in 2x?

Answer 7

yes

Answer 8

ok

Answer 9

x/sqrt(x^2+16)

Answer 10

okay

Answer 11

so it wouldn't be 26

Answer 12

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

Answer 13

okay thx

Answer 14

yw

Answer 15

Thanks for catching the mistake