What is the derivative of 40ln(2x+1) ?

The derivative of 40ln(2x+1) is (80)/(2x+1)

What is the first derivative of 40ln(2x+1) ?

The first derivative of 40ln(2x+1) is (80)/(2x+1)

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derivative of

40 ln (2 x + 1)

\frac{80}{2 x + 1}

Solution steps

\frac{d}{d x} (40 ln (2 x + 1))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 40 \frac{d}{d x} (ln (2 x + 1))

Apply the chain rule:

\frac{1}{2 x + 1} \frac{d}{d x} (2 x + 1)

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

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

= 40 \cdot \frac{1}{2 x + 1} \cdot 2

Simplify

40 \cdot \frac{1}{2 x + 1} \cdot 2 : \frac{80}{2 x + 1}

= \frac{80}{2 x + 1}

n = 0 \sum \infty x^{n} \frac{1}{n}

\frac{d}{d x} (2 x^{3} + \frac{45 x ^{2}}{2} + 21 x + 4)

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

y = \frac{e ^{x} + 1}{e ^{x} - 1}

\frac{\partial}{\partial x} (- \frac{y}{x ^{2}} + 4 x + \frac{1}{y})

What is the derivative of 40ln(2x+1) ?
The derivative of 40ln(2x+1) is (80)/(2x+1)
What is the first derivative of 40ln(2x+1) ?
The first derivative of 40ln(2x+1) is (80)/(2x+1)