What is the derivative of ae^{2x}+be^x ?

The derivative of ae^{2x}+be^x is ae^{2x}*2+be^x

What is the first derivative of ae^{2x}+be^x ?

The first derivative of ae^{2x}+be^x is ae^{2x}*2+be^x

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{\partial}{\partial z} (e^{x} sin (yz))

\int 10 sin^{3} (x) cos^{7} (x) d x

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

\frac{s}{s ^{4} + 1}

\int tan (2 t) d t

What is the derivative of ae^{2x}+be^x ?
The derivative of ae^{2x}+be^x is ae^{2x}*2+be^x
What is the first derivative of ae^{2x}+be^x ?
The first derivative of ae^{2x}+be^x is ae^{2x}*2+be^x