What is the d/(dy)(4e^{2y}(2x^2+4x+2y+1)) ?

The d/(dy)(4e^{2y}(2x^2+4x+2y+1)) is 4(4e^{2y}x^2+8e^{2y}x+4e^{2y}y+4e^{2y})

What is the first d/(dy)(4e^{2y}(2x^2+4x+2y+1)) ?

The first d/(dy)(4e^{2y}(2x^2+4x+2y+1)) is 4(4e^{2y}x^2+8e^{2y}x+4e^{2y}y+4e^{2y})

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\frac{d}{d y} (4 e^{2 y} (2 x^{2} + 4 x + 2 y + 1))

4 (4 e^{2 y} x^{2} + 8 e^{2 y} x + 4 e^{2 y} y + 4 e^{2 y})

Solution steps

\frac{d}{d y} (4 e^{2 y} (2 x^{2} + 4 x + 2 y + 1))

Treat

x

as a constant

Take the constant out:

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

= 4 \frac{d}{d y} (e^{2 y} (2 x^{2} + 4 x + 2 y + 1))

Apply the Product Rule:

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

f = e^{2 y}, g = 2 x^{2} + 4 x + 2 y + 1

= 4 (\frac{d}{d y} (e^{2 y}) (2 x^{2} + 4 x + 2 y + 1) + \frac{d}{d y} (2 x^{2} + 4 x + 2 y + 1) e^{2 y})

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

\frac{d}{d y} (2 x^{2} + 4 x + 2 y + 1) = 2

= 4 (e^{2 y} \cdot 2 (2 x^{2} + 4 x + 2 y + 1) + 2 e^{2 y})

Expand

e^{2 y} \cdot 2 (2 x^{2} + 4 x + 2 y + 1) + 2 e^{2 y} : 4 e^{2 y} x^{2} + 8 e^{2 y} x + 4 e^{2 y} y + 4 e^{2 y}

= 4 (4 e^{2 y} x^{2} + 8 e^{2 y} x + 4 e^{2 y} y + 4 e^{2 y})

\int \frac{( ln ( x ^{3} ) )}{x} d x

x \to \infty lim (\frac{x}{5 x})

\frac{\partial}{\partial x} ((y + c x), (x + k))

x \to \infty lim (\frac{1}{6 x})

\frac{d}{d x} ((- x^{2} + 9)^{2})

What is the d/(dy)(4e^{2y}(2x^2+4x+2y+1)) ?
The d/(dy)(4e^{2y}(2x^2+4x+2y+1)) is 4(4e^{2y}x^2+8e^{2y}x+4e^{2y}y+4e^{2y})
What is the first d/(dy)(4e^{2y}(2x^2+4x+2y+1)) ?
The first d/(dy)(4e^{2y}(2x^2+4x+2y+1)) is 4(4e^{2y}x^2+8e^{2y}x+4e^{2y}y+4e^{2y})