Respuesta :
By the chain rule,
[tex]\dfrac{\partial w}{\partial r}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial r}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial r}[/tex]
and similarly, [tex]\frac{\partial w}{\partial\theta}[/tex] is the same with [tex]r[/tex] replaced with [tex]\theta[/tex].
We have
[tex]\dfrac{\partial w}{\partial x}=y+z[/tex]
[tex]\dfrac{\partial w}{\partial y}=x+z[/tex]
[tex]\dfrac{\partial w}{\partial z}=x+y[/tex]
[tex]\dfrac{\partial x}{\partial r}=\cos\theta[/tex]
[tex]\dfrac{\partial y}{\partial r}=\sin\theta[/tex]
[tex]\dfrac{\partial z}{\partial r}=\theta[/tex]
Putting everything together, we get
[tex]\dfrac{\partial w}{\partial r}=(y+z)\cos\theta+(x+z)\sin\theta+(x+y)\theta[/tex]
Replace [tex]x,y,z[/tex] with their equivalent expressions in terms of [tex]r,\theta[/tex]:
[tex]\dfrac{\partial w}{\partial r}=r(\sin\theta+\theta)\cos\theta+r(\cos\theta+\theta)\sin\theta+r(\cos\theta+\sin\theta)\theta[/tex]
[tex]\dfrac{\partial w}{\partial r}=2r\theta\sin\theta+2r\theta\cos\theta+r\sin(2\theta)[/tex]
When [tex]r=4[/tex] and [tex]\theta=\frac\pi2[/tex], the derivative has a value of
[tex]\dfrac{\partial w}{\partial r}\left(4,\dfrac\pi2\right)=\boxed{4\pi}[/tex]
The same process will yield
[tex]\dfrac{\partial w}{\partial\theta}\left(4,\dfrac\pi2\right)=\boxed{-8\pi}[/tex]
The value of partial derivative are,
[tex]\frac{\partial w }{\partial r}(4,\frac{\pi}{2} )=4\pi[/tex] and [tex]\frac{\partial w }{\partial \theta}(4,\frac{\pi}{2} )=-8\pi[/tex]
Partial derivative :
It is given that,
[tex]w = xy + yz + zx, x = r cos(\theta), y = r sin(\theta), z = r\theta[/tex]
We have to find partial derivative.
[tex]\frac{\partial w }{\partial r}=\frac{\partial w }{\partial x}*\frac{\partial x }{\partial r}+\frac{\partial w }{\partial y}*\frac{\partial y }{\partial r}+\frac{\partial w }{\partial z}*\frac{\partial z }{\partial z}\\\\\frac{\partial w }{\partial r}=(y+z)cos\theta+(x+z)sin\theta+(x+y)\theta\\\\\frac{\partial w }{\partial r}=2r\thetasin\theta+2r\thetacos\theta+rsin(2\theta)[/tex]
When [tex]r=4,\theta=\frac{\pi}{2}[/tex]
[tex]\frac{\partial w }{\partial r}(4,\frac{\pi}{2} )=4\pi[/tex]
Similarly,
[tex]\frac{\partial w }{\partial \theta}(4,\frac{\pi}{2} )=-8\pi[/tex]
Learn more about the chain rule of derivative here:
https://brainly.com/question/11549233
