What is the derivative of Ae^x+Bxe^x ?

The derivative of Ae^x+Bxe^x is Ae^x+B(e^x+e^xx)

What is the first derivative of Ae^x+Bxe^x ?

The first derivative of Ae^x+Bxe^x is Ae^x+B(e^x+e^xx)

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\frac{d}{d x} (A e^{x} + B x e^{x})

A e^{x} + B (e^{x} + e^{x} x)

Solution steps

\frac{d}{d x} (A e^{x} + B x e^{x})

Treat

A, B

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= A e^{x} + B (e^{x} + e^{x} x)

x \to - 3 + lim (x + 1)

\int_{- 1}^{0.5} 1.5 x^{2} d x

\int \frac{x ^{3} - 4 x - 10}{x ^{3} - 3 x ^{2} - 4 x + 12} d x

(x^{2} - 1)^{^{'}}

\int \frac{x}{2 x ^{2} - 1} d x

What is the derivative of Ae^x+Bxe^x ?
The derivative of Ae^x+Bxe^x is Ae^x+B(e^x+e^xx)
What is the first derivative of Ae^x+Bxe^x ?
The first derivative of Ae^x+Bxe^x is Ae^x+B(e^x+e^xx)