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

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

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

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

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

3 e^{x} + \frac{1}{x} + sec^{2} (x)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (tan (x)) = sec^{2} (x)

= 3 e^{x} + \frac{1}{x} + sec^{2} (x)

t \to 0 lim (\frac{7}{t} - \frac{7}{t ^{2} + t})

x \to 2 lim (\frac{x ^{2}}{x - 2})

f (x) = r^{2} - 11 x^{2}

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