What is the derivative of g(x)=\sqrt[4]{x}e^x ?

The derivative of g(x)=\sqrt[4]{x}e^x is (e^x)/(4x^{3/4)}+e^x\sqrt[4]{x}

What is the first derivative of g(x)=\sqrt[4]{x}e^x ?

The first derivative of g(x)=\sqrt[4]{x}e^x is (e^x)/(4x^{3/4)}+e^x\sqrt[4]{x}

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derivative of

Solution steps

Apply the Product Rule:

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

\frac{d}{d x} (e^{x}) = e^{x}

\frac{1}{4 x ^{\frac{3}{4}}} e^{x} = \frac{e ^{x}}{4 x ^{\frac{3}{4}}}

\frac{d y}{d x} = (1 + y^{2}) tan (x)

\frac{d}{d x} (- 6 e^{- 6 x})

x \to - 1 lim (\frac{x ^{3}}{x + 1})

\frac{d}{d x} (\frac{( z x - y ) ^{2}}{x})

\int x^{2} sin (x) cos (x) d x

What is the derivative of g(x)=\sqrt[4]{x}e^x ?
The derivative of g(x)=\sqrt[4]{x}e^x is (e^x)/(4x^{3/4)}+e^x\sqrt[4]{x}
What is the first derivative of g(x)=\sqrt[4]{x}e^x ?
The first derivative of g(x)=\sqrt[4]{x}e^x is (e^x)/(4x^{3/4)}+e^x\sqrt[4]{x}