What is the derivative of c_{1}e^x+c_{2}e^{2x} ?

The derivative of c_{1}e^x+c_{2}e^{2x} is c_{1}e^x+c_{2}e^{2x}*2

What is the first derivative of c_{1}e^x+c_{2}e^{2x} ?

The first derivative of c_{1}e^x+c_{2}e^{2x} is c_{1}e^x+c_{2}e^{2x}*2

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\frac{d}{d x} (c_{1} e^{x} + c_{2} e^{2 x})

c_{1} e^{x} + c_{2} e^{2 x} \cdot 2

Solution steps

\frac{d}{d x} (c_{1} e^{x} + c_{2} e^{2 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (c_{1} e^{x}) + \frac{d}{d x} (c_{2} e^{2 x})

\frac{d}{d x} (c_{1} e^{x}) = c_{1} e^{x}

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

= c_{1} e^{x} + c_{2} e^{2 x} \cdot 2

\frac{d x}{d t} = t^{5} - \frac{2 x}{t}

\int_{0}^{\frac{π}{6}} x sin (x + y) d x

\frac{d}{d x} (ln (1 + 10 x))

x \to 4 + lim (2 x)

What is the derivative of c_{1}e^x+c_{2}e^{2x} ?
The derivative of c_{1}e^x+c_{2}e^{2x} is c_{1}e^x+c_{2}e^{2x}*2
What is the first derivative of c_{1}e^x+c_{2}e^{2x} ?
The first derivative of c_{1}e^x+c_{2}e^{2x} is c_{1}e^x+c_{2}e^{2x}*2