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

1. SithsAndGiggles

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

2. onegirl

so thats the equation? @SithAndGiggles

3. onegirl

@SithsAndGiggles

4. onegirl

@myko

5. myko

the derivative is wrong. The nominator should be just x

6. onegirl

okay u mean in 2x?

7. myko

yes

8. onegirl

ok

9. myko

x/sqrt(x^2+16)

10. onegirl

okay

11. onegirl

so it wouldn't be 26

12. myko

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

13. onegirl

okay thx

14. myko

yw

15. SithsAndGiggles

Thanks for catching the mistake