What is the derivative of (ax+be^x) ?

The derivative of (ax+be^x) is ae^x+e^x(ax+b)

What is the first derivative of (ax+be^x) ?

The first derivative of (ax+be^x) is ae^x+e^x(ax+b)

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = a x + b, g = e^{x}

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

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

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

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

f (x) = x^{2} - x + 10, at x = 3

\int \frac{1}{v \cdot v ^{2} - 1} d v

f (x) = 2 x^{3}, (- 2, - 16)

\int cot (3 x) d x

ln (x x^{2} - 5)

What is the derivative of (ax+be^x) ?
The derivative of (ax+be^x) is ae^x+e^x(ax+b)
What is the first derivative of (ax+be^x) ?
The first derivative of (ax+be^x) is ae^x+e^x(ax+b)