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

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

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

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

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\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}

f (x) = x^{2} - 9, (4, 7)

\frac{\partial}{\partial y} (ln (x^{3} + y^{3}))

6 x + 10, 4 x + 44, x = 15, x = 19

\int_{3.5}^{7} 274000 - 39200 x d x

\frac{d}{d x} (tan (4 x + 4))

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