## mathstina 3 years ago Find the directional derivatives of the following functions at the given points P in the direction of the vectors v. f(x,y,z) =√xyz , P (3,2,6) , v =<-1,-2,2>

1. piscez.in

find the derivatives of the function x, y and z seperately, by keeping the other 2 variables constant in each case. Think of a smart way to keep the other 2 variables constant!

2. mathstina

ok, i tried to work out. is the ans -1?

3. TuringTest

what is you function?$f(x,y,z)=\sqrt{xyz}$or$f(x,y,z)=\sqrt xyz$?

4. mathstina

the first one

5. TuringTest

$D_{\vec v}=\nabla f(3,2,6)\cdot\frac{\vec v}{\|\vec v\|}$I don't think the answer is one but I haven't done it yet...

6. Algebraic!

is there an advanced equation editor available to super users that has cool features like gradient?

7. mathstina

negative one

8. piscez.in

@mathstina @TuringTest is right

9. piscez.in

@mathstina first fing the derivative of the function with respect to x, by substituting the y and x values

10. mathstina

ok

11. TuringTest

oh yeah, I do get -1 :)

12. mathstina

ok thanks a lot!

13. piscez.in

@TuringTest @mathstina no its not -1, atleast as per my calculations

14. TuringTest

I will show my work and you can spot a mistake if you see one...

15. piscez.in

ok

16. TuringTest

$\nabla f=\frac12\langle\sqrt{\frac{yz}x},\sqrt{\frac{xz}y},\sqrt{\frac{xy}z}\rangle\implies\nabla f(3,2,6)=\frac12\langle2,3,1\rangle$$\frac{\vec v}{\|\vec v\|}=\frac13\langle-1,-2,2\rangle$$\nabla f(3,2,6)\cdot\frac{\vec v}{\|\vec v\|}=\frac16(-2-6+2)=-1$

17. mathstina

Great!

18. piscez.in

@TuringTest @mathstina im sorry, your right, i made a mistake by putting all -ves as +ves and vice versa. Bravo you two

19. TuringTest

cheers!

20. mathstina

Its ok. u too got the same ans

21. piscez.in

i got it as 1 previously, now i have -1 :)