What is the derivative of \sqrt[9]{t}-9e^t ?

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

What is the first derivative of \sqrt[9]{t}-9e^t ?

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

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derivative of

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

Solution steps

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) (x + 3 x^{3}))

\int_{- 2}^{2} 4 - x^{2} d x

\int_{0}^{3} 4 - x d x

\frac{d y}{d x} = \frac{x e ^{2 x}}{x + k}

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