What is the derivative of 4\sqrt[3]{x}+2ln(x-3e^x) ?

The derivative of 4\sqrt[3]{x}+2ln(x-3e^x) is (4x+6x^{2/3}-9e^xx^{5/3})/(3x^{5/3)}

What is the first derivative of 4\sqrt[3]{x}+2ln(x-3e^x) ?

The first derivative of 4\sqrt[3]{x}+2ln(x-3e^x) is (4x+6x^{2/3}-9e^xx^{5/3})/(3x^{5/3)}

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\frac{d}{d x} (4 3 x + 2 ln (x) - 3 e^{x})

\frac{4 x + 6 x ^{\frac{2}{3}} - 9 e ^{x} x ^{\frac{5}{3}}}{3 x ^{\frac{5}{3}}}

Solution steps

\frac{d}{d x} (4 3 x + 2 ln (x) - 3 e^{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 3 x) + \frac{d}{d x} (2 ln (x)) - \frac{d}{d x} (3 e^{x})

\frac{d}{d x} (4 3 x) = \frac{4}{3 x ^{\frac{2}{3}}}

\frac{d}{d x} (2 ln (x)) = \frac{2}{x}

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

= \frac{4}{3 x ^{\frac{2}{3}}} + \frac{2}{x} - 3 e^{x}

Simplify

\frac{4}{3 x ^{\frac{2}{3}}} + \frac{2}{x} - 3 e^{x} : \frac{4 x + 6 x ^{\frac{2}{3}} - 9 e ^{x} x ^{\frac{5}{3}}}{3 x ^{\frac{5}{3}}}

= \frac{4 x + 6 x ^{\frac{2}{3}} - 9 e ^{x} x ^{\frac{5}{3}}}{3 x ^{\frac{5}{3}}}

\frac{d}{d x} (\frac{5 x ^{4} + 2 x}{2 x ^{9}})

\int \frac{5 ^{x}}{5 ^{x} - 1} d x

x \to 0 lim (\frac{( 4 + x ) - 16}{x})

f (x) = \frac{5}{6 x ^{5}}

\frac{d}{d x} (9 k \cdot ln (\frac{x + 9}{x - 9}))

What is the derivative of 4\sqrt[3]{x}+2ln(x-3e^x) ?
The derivative of 4\sqrt[3]{x}+2ln(x-3e^x) is (4x+6x^{2/3}-9e^xx^{5/3})/(3x^{5/3)}
What is the first derivative of 4\sqrt[3]{x}+2ln(x-3e^x) ?
The first derivative of 4\sqrt[3]{x}+2ln(x-3e^x) is (4x+6x^{2/3}-9e^xx^{5/3})/(3x^{5/3)}