What is the derivative of f(x)=\sqrt[5]{x}(x+4) ?

The derivative of f(x)=\sqrt[5]{x}(x+4) is (x+4)/(5x^{4/5)}+\sqrt[5]{x}

What is the first derivative of f(x)=\sqrt[5]{x}(x+4) ?

The first derivative of f(x)=\sqrt[5]{x}(x+4) is (x+4)/(5x^{4/5)}+\sqrt[5]{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} (x + 4) = 1

Simplify

\frac{\partial}{\partial x} (- x e^{x} sin (y))

f (x) = x^{2} + 91 x - 92, at x = 2

\int \frac{x ^{3}}{x ^{4} + 2} d x

f (x) = (\frac{x + 7}{x - 7})^{5}

x^{2} - 1, - x + 2, x = 0, x = 1

What is the derivative of f(x)=\sqrt[5]{x}(x+4) ?
The derivative of f(x)=\sqrt[5]{x}(x+4) is (x+4)/(5x^{4/5)}+\sqrt[5]{x}
What is the first derivative of f(x)=\sqrt[5]{x}(x+4) ?
The first derivative of f(x)=\sqrt[5]{x}(x+4) is (x+4)/(5x^{4/5)}+\sqrt[5]{x}