What is the derivative of \sqrt[4]{x}*2^x ?

The derivative of \sqrt[4]{x}*2^x is (2^{x-2})/(x^{3/4)}+ln(2)*2^x\sqrt[4]{x}

What is the first derivative of \sqrt[4]{x}*2^x ?

The first derivative of \sqrt[4]{x}*2^x is (2^{x-2})/(x^{3/4)}+ln(2)*2^x\sqrt[4]{x}

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derivative of

4 x \cdot 2^{x}

\frac{2 ^{x - 2}}{x ^{\frac{3}{4}}} + ln (2) \cdot 2^{x} 4 x

Solution steps

\frac{d}{d x} (4 x \cdot 2^{x})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 4 x, g = 2^{x}

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

\frac{d}{d x} (4 x) = \frac{1}{4 x ^{\frac{3}{4}}}

\frac{d}{d x} (2^{x}) = 2^{x} ln (2)

= \frac{1}{4 x ^{\frac{3}{4}}} \cdot 2^{x} + 2^{x} ln (2) 4 x

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

= \frac{2 ^{x - 2}}{x ^{\frac{3}{4}}} + ln (2) \cdot 2^{x} 4 x

\frac{d y}{d x} = 20 - \frac{y}{6000 + 20 x}

\int_{0}^{π} cos^{2} (2 x) d x

y^{^{''}} + 64 y = 0

\frac{d}{d x} (5 e^{2})

\frac{\partial}{\partial x} (9 x^{4 y})

What is the derivative of \sqrt[4]{x}*2^x ?
The derivative of \sqrt[4]{x}*2^x is (2^{x-2})/(x^{3/4)}+ln(2)*2^x\sqrt[4]{x}
What is the first derivative of \sqrt[4]{x}*2^x ?
The first derivative of \sqrt[4]{x}*2^x is (2^{x-2})/(x^{3/4)}+ln(2)*2^x\sqrt[4]{x}