What is the d/(dy)(2xy^4e^y) ?

The d/(dy)(2xy^4e^y) is 2x(4y^3e^y+e^yy^4)

What is the first d/(dy)(2xy^4e^y) ?

The first d/(dy)(2xy^4e^y) is 2x(4y^3e^y+e^yy^4)

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\frac{d}{d y} (2 x y^{4} e^{y})

2 x (4 y^{3} e^{y} + e^{y} y^{4})

Solution steps

\frac{d}{d y} (2 x y^{4} e^{y})

Treat

x

as a constant

Take the constant out:

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

= 2 x \frac{d}{d y} (y^{4} e^{y})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y^{4}, g = e^{y}

= 2 x (\frac{d}{d y} (y^{4}) e^{y} + \frac{d}{d y} (e^{y}) y^{4})

\frac{d}{d y} (y^{4}) = 4 y^{3}

\frac{d}{d y} (e^{y}) = e^{y}

= 2 x (4 y^{3} e^{y} + e^{y} y^{4})

f (x) 6 x^{4} - 10 x^{3} - 5 x^{2} + 16 x - 45

f (x) = \frac{2}{3 x + 1}, at x = - 1

x \to \frac{π}{3} lim (7 x cos (x))

\frac{d}{d x} (\frac{7}{4 x + 1})

y = e^{3 - x}

What is the d/(dy)(2xy^4e^y) ?
The d/(dy)(2xy^4e^y) is 2x(4y^3e^y+e^yy^4)
What is the first d/(dy)(2xy^4e^y) ?
The first d/(dy)(2xy^4e^y) is 2x(4y^3e^y+e^yy^4)