What is the directional derivative of the function f(x,y,z)= xyz / 1 + (x^2)+(y^2)+(z^2) at the point P(1,1,1) in the direction s=i+4k+j

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oldnick's profile pic

oldnick | (Level 1) Valedictorian

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`(del f)/(del x)=` `(yz(1+y^2+z^2-x^2))/((1+x^2+y^2+z^2)^2)`      `(del f)/(del x) |_(1;1;1)=1/8`

`(del f)/(del y)=` `(xz(1+x^2+z^2-y^2))/((1+x^2+y^2+z^2)^2)`      `(del f)/(del y) |_(1;1;1)=1/8`

`(del f)/(del z)=(xy(1+x^2+y^2-z^2))/((1+x^2+y^2+z^2)^2)`       `(del f)/(del z)|_(1;1;1)=1/8`

So :

`(del f)/(del v)= 1/8 +4xx1/8+1/8=3/4`

oldnick's profile pic

oldnick | (Level 1) Valedictorian

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`f(x,y,z)= (xyz)/(1+x^2+y^2+z^2)`

Set:  `df(x_0,y_0,z_0)` differenzial of function in the point  `P(1;1;1)`

The vector ```v=stackrel(->)(i)+4stackrel(->)(j)+stackrel(->)(k)```  then: 

`(del f)/(del v)= stackrel(->)(df)(x_0;y_0;z_0)) xx stackrel(->)(v)= ` `(del f(x_0;y_0;z_0))/(del x) stackrel (->) (i) xx stackrel(->)(i)+`  `+(del f(x_0;y_0;z_0))/(del y) stackrel(->)(j)xx4stackrel(->)(j)`  `+(del f(x_0;y_0;z_0))/(del z) stackrel(->)(k)xx stackrel(->)(k)=` 

`=(del f(x_0;y_0;z_0))/(del x) + 4 (del f(x_0,y_0;z_0))/(del y)+(del f(x_0;y_0;z_0))/(del z)`

TO BE CONTINUION.....

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pramodpandey | College Teacher | (Level 3) Valedictorian

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Find unit vector parallel to  `s=i+4k+j`

`u=(i+4k+j)/(sqrt(1+4^2+1))`

`=(i+4k+j)/sqrt(18)`        (i)

Wh have  f(x,y,z)= xyz / (1 + (x^2)+(y^2)+(z^2))

grad (f)=`(delf)/(delx)i+(delf)/(dely)j+(delf)/(delz)k`

`(delf)/(delx)={(1+x^2+y^2+z^2)yz-xyz(2x)}/((1+x^2+y^2+z^2)^2)`

`=(1-x^2+y^2+z^2)/(1+x^2+y^2+z^2)`

`((delf)/(delx)) _{(1,1,1)}=2/4`

Similarly `((delf)/(dely))_{(1,1,1)}=2/4`

`((delf)/(delz))_{(1,1,1)}=2/4`

`` grda (f)= `(2/4)i+(2/4)j+(2/4)k`

Thus required direction derivative

`u*grad(f)=((i+4j+k)/sqrt(18))*((i+j+k)/2)`

`=(1+4+1)/(2sqrt(18))`

`=sqrt(2)/2`

Ans.

x_(mn) 

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