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

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

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

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

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derivative of

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

Solution steps

Treat

t

as a constant

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (9 e^{t}) = 0

= \frac{1}{9 x ^{\frac{8}{9}}} - 0

Simplify

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

\int (9 x^{2} - 8 x + 3) d x

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\int \frac{4}{x ^{2} - 4 x + 20} d x

x, x, 0, 2

\int (x^{\frac{3}{2}} + 5)^{6} x d x

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