What is the derivative of y=ln(e^x+xe^x) ?

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

What is the first derivative of y=ln(e^x+xe^x) ?

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

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derivative of

y = ln (e^{x} + x e^{x})

\frac{2 + x}{1 + x}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{2 + x}{1 + x}

\frac{d}{d x} (3 (9 x - 4)^{4})

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

y^{^{'}} + 2 \cdot y = t

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

\frac{\partial}{\partial x} (e^{x} ln (x y))

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