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

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

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

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

8 x^{4 y} ln (x)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

x^{4 y} = e^{4 y l n (x)}

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

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (4 y ln (x)) = 4 ln (x)

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

Simplify

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

= 8 x^{4 y} ln (x)

\frac{d}{d x} (9 tan (x) - 9 cot (x))

\frac{d}{d x} (x (17 - 2 x) (33 - 2 x))

x \to 0 lim (2 x^{2} + 4 x - 2)

1 2^{^{'}}

x \to 1 lim (\frac{2}{x ^{2} - 1})

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