Answer
$\frac{∂w}{∂ρ} = 2ρ(4sin^2 φ +cos^2 φ)$
$\frac{∂w}{∂φ} = 6 ρ^2sin φcos φ $
$\frac{∂w}{∂θ} = 0$
Work Step by Step
Given that,
$w = 4x^2 + 4y^2 + z^2,
x = ρ sin φ cos θ,
y = ρ sin φ sin θ,
z = ρ cos φ$
Now let's take $x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ$ in $w$.
So we get,
$w = 4(ρ sin φ cos θ)^2+4(ρ sin φ sin θ)^2+(ρ cos φ)^2$
$w = 4ρ^2 sin^2 φ cos^2 θ+4ρ^2 sin^2 φ sin^2 θ+ρ^2 cos^2 φ$
$w = ( cos^2 θ+ sin^2 θ)4ρ^2 sin^2 φ+ρ^2 cos^2 φ$
$w = 4ρ^2 sin^2 φ+ρ^2 cos^2 φ$ [As, $cos^2 θ+sin^2 θ =1$]
Now calculate the partial derivatives $\frac{∂w}{∂ρ}, \frac{∂w}{∂φ}$ and $\frac{∂w}{∂θ}.$
$\frac{∂w}{∂ρ} = \frac{∂}{∂ρ}[4ρ^2 sin^2 φ+ρ^2 cos^2 φ]$
$\frac{∂w}{∂ρ} = 4 sin^2 φ \frac{∂}{∂ρ}[ρ^2]+cos^2 φ\frac{∂}{∂ρ}[ρ^2]$
$\frac{∂w}{∂ρ} = 4.2 ρsin^2 φ +2ρcos^2 φ$
$\frac{∂w}{∂ρ} = 2ρ(4sin^2 φ +cos^2 φ)$
$\frac{∂w}{∂φ} = \frac{∂}{∂φ}[4ρ^2 sin^2 φ+ρ^2 cos^2 φ]$
$\frac{∂w}{∂φ} = 4 ρ^2 \frac{∂}{∂φ}[sin^2 φ]+ρ^2\frac{∂}{∂φ}[cos^2 φ]$
$\frac{∂w}{∂φ} = 4 ρ^2.2sin φ\frac{∂}{∂φ}[sin φ] +ρ^2.2cos φ\frac{∂}{∂φ}[cos φ]$
$\frac{∂w}{∂φ} = 8 ρ^2sin φcos φ +2ρ^2cos φ(- sin φ)$
$\frac{∂w}{∂φ} = 8 ρ^2sin φcos φ -2ρ^2sin φcos φ$
$\frac{∂w}{∂φ} = 6 ρ^2sin φcos φ $
$\frac{∂w}{∂θ} = \frac{∂}{∂θ}[4ρ^2 sin^2 φ+ρ^2 cos^2 φ]$
$\frac{∂w}{∂θ} = 0$
