What is (\partial)/(\partial x)(1/4 2^xy^3) ?

The answer to (\partial)/(\partial x)(1/4 2^xy^3) is ln(2)*2^{x-2}y^3

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(\partial)/(\partial x)(1/4 2^xy^3)

\frac{\partial}{\partial x} (\frac{1}{4} 2^{x} y^{3})

ln (2) \cdot 2^{x - 2} y^{3}

Solution steps

\frac{\partial}{\partial x} (\frac{1}{4} \cdot 2^{x} y^{3})

Simplify

\frac{1}{4} \cdot 2^{x} y^{3} : 2^{x - 2} y^{3}

= \frac{\partial}{\partial x} (2^{x - 2} y^{3})

Treat

y

as a constant

Take the constant out:

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

= y^{3} \frac{\partial}{\partial x} (2^{x - 2})

Apply exponent rule:

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

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

= y^{3} \frac{\partial}{\partial x} (e^{(x - 2) l n (2)})

Apply the chain rule:

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

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

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

= y^{3} e^{(x - 2) l n (2)} ln (2)

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

= ln (2) \cdot 2^{x - 2} y^{3}

\int \frac{1}{16 + x ^{2}} d x

\frac{\partial}{\partial y} (8 x y - \frac{1}{4} (x + y)^{4})

x \to x lim (sin (x))

y = x^{2}, y = 9 x, y = - 1

\frac{d}{d x} (- x e^{- x} - e^{- x} + 1)

What is (\partial)/(\partial x)(1/4 2^xy^3) ?
The answer to (\partial)/(\partial x)(1/4 2^xy^3) is ln(2)*2^{x-2}y^3