What is the derivative of f(x)=\sqrt[8]{x}-8e^x ?

The derivative of f(x)=\sqrt[8]{x}-8e^x is 1/(8x^{7/8)}-8e^x

What is the first derivative of f(x)=\sqrt[8]{x}-8e^x ?

The first derivative of f(x)=\sqrt[8]{x}-8e^x is 1/(8x^{7/8)}-8e^x

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derivative of

\frac{1}{8 x ^{\frac{7}{8}}} - 8 e^{x}

Solution steps

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d x} (8 e^{x}) = 8 e^{x}

= \frac{1}{8 x ^{\frac{7}{8}}} - 8 e^{x}

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

\frac{d}{d x} (\frac{3 x ^{2} - 3}{- 2 x})

\frac{\partial}{\partial x} (8 x^{2})

3 y^{^{''}} - 5 y^{^{'}} - 2 y = x^{3} - 2 x

x \to - 2 lim (x^{3} - 2 x^{2} + 1)

What is the derivative of f(x)=\sqrt[8]{x}-8e^x ?
The derivative of f(x)=\sqrt[8]{x}-8e^x is 1/(8x^{7/8)}-8e^x
What is the first derivative of f(x)=\sqrt[8]{x}-8e^x ?
The first derivative of f(x)=\sqrt[8]{x}-8e^x is 1/(8x^{7/8)}-8e^x