What is the derivative of a(e^{bx}+be^{bx}x) ?

The derivative of a(e^{bx}+be^{bx}x) is a(b^2e^{bx}x+2be^{bx})

What is the first derivative of a(e^{bx}+be^{bx}x) ?

The first derivative of a(e^{bx}+be^{bx}x) is a(b^2e^{bx}x+2be^{bx})

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\frac{d}{d x} (a (e^{b x} + b e^{b x} x))

a (b^{2} e^{b x} x + 2 b e^{b x})

Solution steps

\frac{d}{d x} (a (e^{b x} + b e^{b x} x))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

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

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

= a (e^{b x} b + b (b e^{b x} x + e^{b x}))

Expand

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

= a (b^{2} e^{b x} x + 2 b e^{b x})

\frac{\partial}{\partial x} (x (4 - x^{2}))

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} e^{x}

y^{^{'}} - \frac{y}{x} = x cos (x)

\frac{d}{d x} (\frac{1}{4} x^{4} - 2 x^{2})

d x + e^{8 x} d y = 0

What is the derivative of a(e^{bx}+be^{bx}x) ?
The derivative of a(e^{bx}+be^{bx}x) is a(b^2e^{bx}x+2be^{bx})
What is the first derivative of a(e^{bx}+be^{bx}x) ?
The first derivative of a(e^{bx}+be^{bx}x) is a(b^2e^{bx}x+2be^{bx})