What is the d/(dy)(({g}(y,x))(-e^x)) ?

The d/(dy)(({g}(y,x))(-e^x)) is -e^x d/(dy)({g}(y,x))

What is the first d/(dy)(({g}(y,x))(-e^x)) ?

The first d/(dy)(({g}(y,x))(-e^x)) is -e^x d/(dy)({g}(y,x))

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\frac{d}{d y} ((g (y, x)) (- e^{x}))

- e^{x} \frac{d}{d y} (g (y, x))

Solution steps

\frac{d}{d y} ((g (y, x)) (- e^{x}))

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - e^{x} \frac{d}{d y} (g (y, x))

(x) (x + 1) (x + 2) (x + 3)

4 9^{3}

x + 2 (2 x)

x^{2 - 6 x + 5}

4 x^{22} - 36

What is the d/(dy)(({g}(y,x))(-e^x)) ?
The d/(dy)(({g}(y,x))(-e^x)) is -e^x d/(dy)({g}(y,x))
What is the first d/(dy)(({g}(y,x))(-e^x)) ?
The first d/(dy)(({g}(y,x))(-e^x)) is -e^x d/(dy)({g}(y,x))