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Popular >

d/(dy)(2xy^4e^y+2xy^3+y)

Solution

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

Solution

2 x (4 y^{3} e^{y} + e^{y} y^{4}) + 6 x y^{2} + 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d y} (2 x y^{3}) = 6 x y^{2}

\frac{d y}{d y} = 1

= 2 x (4 y^{3} e^{y} + e^{y} y^{4}) + 6 x y^{2} + 1

Popular Examples

y^{''}+4y^'+6y=cos(x),y(0)=0,y^'(0)=2

y^{^{''}} + 4 y^{^{'}} + 6 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2

sum from n=1 to infinity of (x^n)/(2n-1)

n = 1 \sum \infty \frac{x ^{n}}{2 n - 1}

(\partial)/(\partial x)(e^{xz})

\frac{\partial}{\partial x} (e^{x z})

limit as x approaches 0+of (1-4x)^{1/x}

x \to 0 + lim ((1 - 4 x)^{\frac{1}{x}})

integral of 6x(3x^2+4)^4

\int 6 x (3 x^{2} + 4)^{4} d x

Frequently Asked Questions (FAQ)

What is the d/(dy)(2xy^4e^y+2xy^3+y) ?
The d/(dy)(2xy^4e^y+2xy^3+y) is 2x(4y^3e^y+e^yy^4)+6xy^2+1
What is the first d/(dy)(2xy^4e^y+2xy^3+y) ?
The first d/(dy)(2xy^4e^y+2xy^3+y) is 2x(4y^3e^y+e^yy^4)+6xy^2+1