Answer
$\frac{∂w}{∂x} = \frac{3}{2}( 3x^2 + 2)(\sqrt(x-1))[39x^3-30x^2 + 10x-4 ]$
Work Step by Step
Given that,
$w = 3xy^2z^3,$
$ y = 3x^2 + 2, $
$z = \sqrt(x-1)$
Now calculate the partial derivatives $\frac{∂y}{∂x}$ and $\frac{∂z}{∂x}.$
$\frac{∂y}{∂x} = \frac{∂}{∂x}[3x^2 + 2]$
$\frac{∂y}{∂x} = 3.2x$
$\frac{∂y}{∂x} = 6x$
$\frac{∂z}{∂x} = \frac{∂}{∂x}[ \sqrt(x-1)]$
$\frac{∂z}{∂x} = \frac{1}{2\sqrt(x-1)}$
Now again calculate the partial derivatives $\frac{∂w}{∂x},\frac{∂w}{∂y}$ and $\frac{∂w}{∂z}.$
$\frac{∂w}{∂x} = \frac{∂}{∂x}[3xy^2z^3]$
$\frac{∂w}{∂x} = 3y^2z^3$ [Considering $y$ and $z$ as constants]
$\frac{∂w}{∂y} = \frac{∂}{∂y}[3xy^2z^3]$
$\frac{∂w}{∂y} = 3xz^3.2y$ [Considering $x$ and $z$ as constants]
$\frac{∂w}{∂y} = 6xyz^3$
$\frac{∂w}{∂z} = \frac{∂}{∂z}[3xy^2z^3]$
$\frac{∂w}{∂z} = 3xy^2.3z^2$ [Considering $x$ and $y$ as constants]
$\frac{∂w}{∂z} = 9xy^2z^2$
According to the chain rule for derivatives
$\frac{∂w}{∂x} = \frac{∂w}{∂x}+\frac{∂w}{∂y}\frac{∂y}{∂x}+\frac{∂w}{∂z}\frac{∂z}{∂x}$
$\frac{∂w}{∂x} = [3y^2z^3] + [6xyz^3][6x] + [9xy^2z^2][\frac{1}{2\sqrt(x-1)}]$
$\frac{∂w}{∂x} = [3y^2z^3] + [36x^2yz^3] + [9xy^2z^2][\frac{1}{2\sqrt(x-1)}]$
$\frac{∂w}{∂x} = [3y^2z^3] + [36x^2yz^3] + [9xy^2z^2][\frac{1}{2z}]$ [As $ z=\sqrt(x-1)]$
$\frac{∂w}{∂x} = [3y^2z^3] + [36x^2yz^3] + [9xy^2z][\frac{1}{2}]$
$\frac{∂w}{∂x} = [\frac{3}{2}yz][2yz^2 + 24x^2z^2 + 3xy]$ --(i)
Now if we put the values of $ y$ and $z$ in (i) we get,
$\frac{∂w}{∂x} = [\frac{3}{2}( 3x^2 + 2)(\sqrt(x-1))][2( 3x^2 + 2)(\sqrt(x-1))^2 + 24x^2(\sqrt(x-1))^2 + 3x( 3x^2 + 2)]$
$\frac{∂w}{∂x} = \frac{3}{2}( 3x^2 + 2)(\sqrt(x-1))[2( 3x^2 + 2)(x-1) + 24x^2(x-1) + 3x( 3x^2 + 2)]$
$\frac{∂w}{∂x} = \frac{3}{2}( 3x^2 + 2)(\sqrt(x-1))[( 6x^2 + 4)(x-1) + 24x^3-24x^2 + 9x^3 + 6x]$
$\frac{∂w}{∂x} = \frac{3}{2}( 3x^2 + 2)(\sqrt(x-1))[6x^3-6x^2 + 4x-4 + 24x^3-24x^2 + 9x^3 + 6x]$
$\frac{∂w}{∂x} = \frac{3}{2}( 3x^2 + 2)(\sqrt(x-1))[39x^3-30x^2 + 10x-4 ]$
