What is the derivative of ln(x^2e^x) ?

The derivative of ln(x^2e^x) is 2/x+1

What is the first derivative of ln(x^2e^x) ?

The first derivative of ln(x^2e^x) is 2/x+1

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\frac{d}{d x} (ln (x^{2} e^{x}))

\frac{2}{x} + 1

Solution steps

\frac{d}{d x} (ln (x^{2} e^{x}))

Simplify

ln (x^{2} e^{x}) : ln (x^{2}) + x

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d x}{d x} = 1

= \frac{2}{x} + 1

n = 0 \sum \infty \frac{1}{( n ^{3} - n )}

y^{^{'}} + 8 x - 1 y = x^{8}, y (1) = 9

p p

\int \frac{( 3 x + 1 ) ^{2}}{244} d x

\frac{d}{d θ} (sin (θ) + θ cos (θ))

What is the derivative of ln(x^2e^x) ?
The derivative of ln(x^2e^x) is 2/x+1
What is the first derivative of ln(x^2e^x) ?
The first derivative of ln(x^2e^x) is 2/x+1