What is the derivative of (ax+be^{2x}) ?

The derivative of (ax+be^{2x}) is ae^{2x}+e^{2x}*2(ax+b)

What is the first derivative of (ax+be^{2x}) ?

The first derivative of (ax+be^{2x}) is ae^{2x}+e^{2x}*2(ax+b)

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

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

Solution steps

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

Apply the Product Rule:

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

f = a x + b, g = e^{2 x}

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

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

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

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

\int \frac{x + 4}{x ^{2} + 9} d x

\frac{\partial}{\partial y} (x y - 2 y)

f (x) g (x) h (x)

2 x^{2} + 8

f (x) = 6 e^{x} + 5 x - ln (x)

What is the derivative of (ax+be^{2x}) ?
The derivative of (ax+be^{2x}) is ae^{2x}+e^{2x}*2(ax+b)
What is the first derivative of (ax+be^{2x}) ?
The first derivative of (ax+be^{2x}) is ae^{2x}+e^{2x}*2(ax+b)