What is the derivative of f(x)=\sqrt[9]{x}-9e^x ?

The derivative of f(x)=\sqrt[9]{x}-9e^x is 1/(9x^{8/9)}-9e^x

What is the first derivative of f(x)=\sqrt[9]{x}-9e^x ?

The first derivative of f(x)=\sqrt[9]{x}-9e^x is 1/(9x^{8/9)}-9e^x

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derivative of

\frac{1}{9 x ^{\frac{8}{9}}} - 9 e^{x}

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (9 e^{x}) = 9 e^{x}

= \frac{1}{9 x ^{\frac{8}{9}}} - 9 e^{x}

x \to \frac{3 π}{2} lim (ln (1 + sin (x)))

\frac{d}{d y} (2 x + 3 y)

\int_{4}^{8} \frac{x}{x ^{2} - 1} d x

\frac{d}{d x} (3 x e^{x})

y = e^{8 x} cos (11 x)

What is the derivative of f(x)=\sqrt[9]{x}-9e^x ?
The derivative of f(x)=\sqrt[9]{x}-9e^x is 1/(9x^{8/9)}-9e^x
What is the first derivative of f(x)=\sqrt[9]{x}-9e^x ?
The first derivative of f(x)=\sqrt[9]{x}-9e^x is 1/(9x^{8/9)}-9e^x