What is the derivative of f(x)=ln(6x^3+e^x) ?

The derivative of f(x)=ln(6x^3+e^x) is (18x^2+e^x)/(6x^3+e^x)

What is the first derivative of f(x)=ln(6x^3+e^x) ?

The first derivative of f(x)=ln(6x^3+e^x) is (18x^2+e^x)/(6x^3+e^x)

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derivative of

f (x) = ln (6 x^{3} + e^{x})

\frac{18 x ^{2} + e ^{x}}{6 x ^{3} + e ^{x}}

Solution steps

\frac{d}{d x} (ln (6 x^{3} + e^{x}))

Apply the chain rule:

\frac{1}{6 x ^{3} + e ^{x}} \frac{d}{d x} (6 x^{3} + e^{x})

= \frac{1}{6 x ^{3} + e ^{x}} \frac{d}{d x} (6 x^{3} + e^{x})

\frac{d}{d x} (6 x^{3} + e^{x}) = 18 x^{2} + e^{x}

= \frac{1}{6 x ^{3} + e ^{x}} (18 x^{2} + e^{x})

Simplify

\frac{1}{6 x ^{3} + e ^{x}} (18 x^{2} + e^{x}) : \frac{18 x ^{2} + e ^{x}}{6 x ^{3} + e ^{x}}

= \frac{18 x ^{2} + e ^{x}}{6 x ^{3} + e ^{x}}

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

y^{^{'}} = \frac{y - x}{x}

\frac{\partial}{\partial x} (4 x^{5} y)

\frac{\partial}{\partial x} (\frac{x ^{2} - 1}{x ^{2} + x + 1})

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What is the derivative of f(x)=ln(6x^3+e^x) ?
The derivative of f(x)=ln(6x^3+e^x) is (18x^2+e^x)/(6x^3+e^x)
What is the first derivative of f(x)=ln(6x^3+e^x) ?
The first derivative of f(x)=ln(6x^3+e^x) is (18x^2+e^x)/(6x^3+e^x)