What is (\partial)/(\partial y)(4x^{3y}) ?

The answer to (\partial)/(\partial y)(4x^{3y}) is 12x^{3y}ln(x)

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(\partial)/(\partial y)(4x^{3y})

\frac{\partial}{\partial y} (4 x^{3 y})

12 x^{3 y} ln (x)

Solution steps

\frac{\partial}{\partial y} (4 x^{3 y})

Treat

x

as a constant

Take the constant out:

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

= 4 \frac{\partial}{\partial y} (x^{3 y})

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{3 y} = e^{3 y l n (x)}

= 4 \frac{\partial}{\partial y} (e^{3 y l n (x)})

Apply the chain rule:

e^{3 y l n (x)} \frac{\partial}{\partial y} (3 y ln (x))

= e^{3 y l n (x)} \frac{\partial}{\partial y} (3 y ln (x))

\frac{\partial}{\partial y} (3 y ln (x)) = 3 ln (x)

= 4 e^{3 y l n (x)} \cdot 3 ln (x)

Simplify

4 e^{3 y l n (x)} \cdot 3 ln (x) : 12 x^{3 y} ln (x)

= 12 x^{3 y} ln (x)

3 y^{^{'}} + 4 y = e^{- x}

tan (5 - sin (2 t))

\frac{d y}{d x} + 3 x^{2} y = 2 x e^{- x^{3}}

y^{^{'}} + 2 x^{- 1} y = x^{- 1} sin (2 x)

\int_{- 2}^{1} x (4 - x^{2}) - (x + 2) d x

What is (\partial)/(\partial y)(4x^{3y}) ?
The answer to (\partial)/(\partial y)(4x^{3y}) is 12x^{3y}ln(x)