What is the derivative of y=ln(3x)+x ?

The derivative of y=ln(3x)+x is 1/x+1

What is the first derivative of y=ln(3x)+x ?

The first derivative of y=ln(3x)+x is 1/x+1

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derivative of

y = ln (3 x) + x

\frac{1}{x} + 1

Solution steps

\frac{d}{d x} (ln (3 x) + x)

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (ln (3 x)) = \frac{1}{x}

\frac{d x}{d x} = 1

= \frac{1}{x} + 1

\int \frac{sin ( 2 x )}{26 + cos ^{2} ( x )} d x

\int \frac{3 x ^{2} + x + 4}{x ^{3} + x} d x

n = 1 \sum \infty \frac{1}{n ^{2} + 3 n}

f (x) = 270

y = x^{3} - 4 x^{2} + 3 x, y = x^{2} - x

What is the derivative of y=ln(3x)+x ?
The derivative of y=ln(3x)+x is 1/x+1
What is the first derivative of y=ln(3x)+x ?
The first derivative of y=ln(3x)+x is 1/x+1