## anonymous 3 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. 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: $y-\sqrt{185}=\frac{26}{\sqrt{185}}(x-13)$

2. anonymous

so thats the equation? @SithAndGiggles

3. anonymous

@SithsAndGiggles

4. anonymous

@myko

5. anonymous

the derivative is wrong. The nominator should be just x

6. anonymous

okay u mean in 2x?

7. anonymous

yes

8. anonymous

ok

9. anonymous

x/sqrt(x^2+16)

10. anonymous

okay

11. anonymous

so it wouldn't be 26

12. anonymous

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

13. anonymous

okay thx

14. anonymous

yw

15. anonymous

Thanks for catching the mistake