What is (\partial)/(\partial y)(4x^{2y}) ?

The answer to (\partial)/(\partial y)(4x^{2y}) is 8x^{2y}ln(x)

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(\partial)/(\partial y)(4x^{2y})

\frac{\partial}{\partial y} (4 x^{2 y})

8 x^{2 y} ln (x)

Solution steps

\frac{\partial}{\partial y} (4 x^{2 y})

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

x^{2 y} = e^{2 y l n (x)}

= 4 \frac{\partial}{\partial y} (e^{2 y l n (x)})

Apply the chain rule:

e^{2 y l n (x)} \frac{\partial}{\partial y} (2 y ln (x))

= e^{2 y l n (x)} \frac{\partial}{\partial y} (2 y ln (x))

\frac{\partial}{\partial y} (2 y ln (x)) = 2 ln (x)

= 4 e^{2 y l n (x)} \cdot 2 ln (x)

Simplify

4 e^{2 y l n (x)} \cdot 2 ln (x) : 8 x^{2 y} ln (x)

= 8 x^{2 y} ln (x)

\frac{d ^{2} y}{d x ^{2}} = x

f (x) = \frac{x}{x ^{2} + 1}

\frac{d}{d x} (\frac{cot ( x )}{1 + cot ( x )})

\frac{d y}{d x}, y - 5 x y^{3} + 8 = 5 sin (y^{3} + 2 x^{2})

\frac{\partial}{\partial x} (x (y + 1))

What is (\partial)/(\partial y)(4x^{2y}) ?
The answer to (\partial)/(\partial y)(4x^{2y}) is 8x^{2y}ln(x)