What is the derivative of 4*(sqrt(x^2+3x)) ?

The derivative of 4*(sqrt(x^2+3x)) is (2(2x+3))/(sqrt(x^2+3x))

What is the first derivative of 4*(sqrt(x^2+3x)) ?

The first derivative of 4*(sqrt(x^2+3x)) is (2(2x+3))/(sqrt(x^2+3x))

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

\frac{2 ( 2 x + 3 )}{x ^{2} + 3 x}

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= 4 \cdot \frac{1}{2 x ^{2} + 3 x} (2 x + 3)

Simplify

4 \cdot \frac{1}{2 x ^{2} + 3 x} (2 x + 3) : \frac{2 ( 2 x + 3 )}{x ^{2} + 3 x}

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

\frac{d}{d x} (4 sec^{2} (2 x))

\int \frac{( 1 - t ) ^{2}}{t} d t

\frac{d}{d l} (2 π \frac{l}{9.8})

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

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

What is the derivative of 4*(sqrt(x^2+3x)) ?
The derivative of 4*(sqrt(x^2+3x)) is (2(2x+3))/(sqrt(x^2+3x))
What is the first derivative of 4*(sqrt(x^2+3x)) ?
The first derivative of 4*(sqrt(x^2+3x)) is (2(2x+3))/(sqrt(x^2+3x))