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${:[x^(3),-x(y-z)^(2)],[y^(3),-y(z-x)^(9)],[z^(3),-z(x-y)^(9)]:}$

$\begin{array}{ll}x^{3} & -x(y-z)^{2} \\ y^{3} & -y(z-x)^{9} \\ z^{3} & -z(x-y)^{9}\end{array}$

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Find derivatives of sine and cosine functions

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Q.

\begin{array}{ll}x^{3} & -x(y-z)^{2} \\ y^{3} & -y(z-x)^{9} \\ z^{3} & -z(x-y)^{9}\end{array}

Differentiate first row: Differentiate the first row with respect to $x$ . $\frac{d}{dx}[x^{3}, -x(y-z)^{2}] = [3x^{2}, -1\cdot(y-z)^{2} - x\cdot2(y-z)\cdot\frac{d}{dx}(y-z)]$
Differentiate second row: Since $y$ and $z$ are constants with respect to $x$ , $\frac{d}{dx}(y-z) = 0$ . $\newline$ \frac{d}{dx}\left[ x^{\(3\)}, -x(y-z)^{\(2\)} \right] = \left[ \(3x^{ $2$ }, $-1$ \cdot(y-z)^{ $2$ } - x\cdot $2$ (y-z)\cdot $0$ \right] = \left[ $3$ x^{ $2$ }, -(y-z)^{ $2$ } \right]
Differentiate third row: Differentiate the second row with respect to $x$ . $\frac{d}{dx}\left[ y^{3}, -y(z-x)^{9} \right] = \left[ 0, -y\cdot 9(z-x)^{8}\cdot\frac{d}{dx}(z-x) \right]$
Differentiate third row: Differentiate the second row with respect to $x$ .
$\frac{d}{dx}[y^{3}, -y(z-x)^{9}] = [0, -y\cdot 9(z-x)^{8}\cdot\frac{d}{dx}(z-x)]$ Since $y$ and $z$ are constants with respect to $x$ , $\frac{d}{dx}(z-x) = -1$ .
$\frac{d}{dx}[y^{3}, -y(z-x)^{9}] = [0, -y\cdot 9(z-x)^{8}\cdot(-1)] = [0, 9y(z-x)^{8}]$
Differentiate third row: Differentiate the second row with respect to $x$ .
$\frac{d}{dx}[y^{3}, -y(z-x)^{9}] = [0, -y\cdot 9(z-x)^{8}\cdot\frac{d}{dx}(z-x)]$ Since $y$ and $z$ are constants with respect to $x$ , $\frac{d}{dx}(z-x) = -1$ .
$\frac{d}{dx}[y^{3}, -y(z-x)^{9}] = [0, -y\cdot 9(z-x)^{8}\cdot(-1)] = [0, 9y(z-x)^{8}]$ Differentiate the third row with respect to $x$ .
$\frac{d}{dx}[z^{3}, -z(x-y)^{9}] = [0, -z\cdot 9(x-y)^{8}\cdot\frac{d}{dx}(x-y)]$
Differentiate third row: Differentiate the second row with respect to $x$ . $\frac{d}{dx}[y^{3}, -y(z-x)^{9}] = [0, -y\cdot 9(z-x)^{8}\cdot\frac{d}{dx}(z-x)]$ Since $y$ and $z$ are constants with respect to $x$ , $\frac{d}{dx}(z-x) = -1$ . $\frac{d}{dx}[y^{3}, -y(z-x)^{9}] = [0, -y\cdot 9(z-x)^{8}\cdot(-1)] = [0, 9y(z-x)^{8}]$ Differentiate the third row with respect to $x$ . $\frac{d}{dx}[z^{3}, -z(x-y)^{9}] = [0, -z\cdot 9(x-y)^{8}\cdot\frac{d}{dx}(x-y)]$ Since $y$ and $z$ are constants with respect to $x$ , $\frac{d}{dx}(x-y) = 1$ . $\frac{d}{dx}[z^{3}, -z(x-y)^{9}] = [0, -z\cdot 9(x-y)^{8}\cdot 1] = [0, -9z(x-y)^{8}]$

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