What is the derivative of ln(xe^{sqrt(x)}+9) ?

The derivative of ln(xe^{sqrt(x)}+9) is (e^{sqrt(x)}sqrt(x)+2e^{sqrt(x)})/(2(e^{sqrt(x))x+9)}

What is the first derivative of ln(xe^{sqrt(x)}+9) ?

The first derivative of ln(xe^{sqrt(x)}+9) is (e^{sqrt(x)}sqrt(x)+2e^{sqrt(x)})/(2(e^{sqrt(x))x+9)}

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derivative of

ln (x e^{x} + 9)

\frac{e ^{x} x + 2 e ^{x}}{2 ( e ^{x} x + 9 )}

Solution steps

\frac{d}{d x} (ln (x e^{x} + 9))

Apply the chain rule:

\frac{1}{x e ^{x} + 9} \frac{d}{d x} (x e^{x} + 9)

= \frac{1}{x e ^{x} + 9} \frac{d}{d x} (x e^{x} + 9)

\frac{d}{d x} (x e^{x} + 9) = \frac{e ^{x} x}{2} + e^{x}

= \frac{1}{x e ^{x} + 9} (\frac{e ^{x} x}{2} + e^{x})

Simplify

\frac{1}{x e ^{x} + 9} (\frac{e ^{x} x}{2} + e^{x}) : \frac{e ^{x} x + 2 e ^{x}}{2 ( e ^{x} x + 9 )}

= \frac{e ^{x} x + 2 e ^{x}}{2 ( e ^{x} x + 9 )}

f (x) = \frac{1}{x ^{2}}, at x = - 1

x \to \infty lim (ln (5))

\int ln (6 x^{4}) d x

\int_{0}^{x} \frac{1}{x} d x

\frac{d}{d x} (9 cos^{2} (x))

What is the derivative of ln(xe^{sqrt(x)}+9) ?
The derivative of ln(xe^{sqrt(x)}+9) is (e^{sqrt(x)}sqrt(x)+2e^{sqrt(x)})/(2(e^{sqrt(x))x+9)}
What is the first derivative of ln(xe^{sqrt(x)}+9) ?
The first derivative of ln(xe^{sqrt(x)}+9) is (e^{sqrt(x)}sqrt(x)+2e^{sqrt(x)})/(2(e^{sqrt(x))x+9)}