What is the derivative of 2e^x+3x-ln(x) ?

The derivative of 2e^x+3x-ln(x) is 2e^x+3-1/x

What is the first derivative of 2e^x+3x-ln(x) ?

The first derivative of 2e^x+3x-ln(x) is 2e^x+3-1/x

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\frac{d}{d x} (2 e^{x} + 3 x - ln (x))

2 e^{x} + 3 - \frac{1}{x}

Solution steps

\frac{d}{d x} (2 e^{x} + 3 x - ln (x))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 e^{x}) + \frac{d}{d x} (3 x) - \frac{d}{d x} (ln (x))

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

\frac{d}{d x} (3 x) = 3

\frac{d}{d x} (ln (x)) = \frac{1}{x}

= 2 e^{x} + 3 - \frac{1}{x}

\int - \frac{1}{2} sin (2 x) d x

x \to 80 lim (82 - x)

\int 2 sin (π x) d x

x^{4} - 6 x^{3}

x \to - 2 - lim (2 x + 1)

What is the derivative of 2e^x+3x-ln(x) ?
The derivative of 2e^x+3x-ln(x) is 2e^x+3-1/x
What is the first derivative of 2e^x+3x-ln(x) ?
The first derivative of 2e^x+3x-ln(x) is 2e^x+3-1/x