differentiate sqrt(4x+1) — step-by-step solution

Let's perform the differentiation:

\frac{d}{d x} (\sqrt{4 x + 1})

Apply the differentiation rule:

\frac{d}{d x} (\sqrt{f}) = \frac{d}{d x} (f) \cdot \frac{1}{2 \sqrt{f}}

with f =

4 x + 1

ⓘ The derivative of

\sqrt{x}

\frac{1}{2 \sqrt{x}}

\frac{d}{d x} (\sqrt{4 x + 1})

\frac{d}{d x} (4 x + 1) \cdot \frac{1}{2 \cdot \sqrt{4 x + 1}}

\frac{d}{d x} (4 x + 1) \cdot \frac{1}{2 \cdot \sqrt{4 x + 1}}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{d}{d x} (4 x + 1) \cdot \frac{1}{2 \cdot \sqrt{4 x + 1}}

\frac{\frac{d}{d x} (4 x + 1) \cdot 1}{2 \cdot \sqrt{4 x + 1}}

\frac{\frac{d}{d x} (4 x + 1)}{2 \cdot \sqrt{4 x + 1}}

■The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (4 x + 1)

\frac{d}{d x} (4 x) + 0

\frac{\frac{d}{d x} (4 x) + 0}{2 \cdot \sqrt{4 x + 1}}

ⓘ Adding 0 preserves the original value:

\frac{d}{d x} (4 x) + 0

\frac{d}{d x} (4 x)

■The derivative of a constant multiple keeps that factor: that of

k u

k \cdot u'

\frac{d}{d x} (4 x)

4

\frac{4}{2 \cdot \sqrt{4 x + 1}}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same nonzero number:

\frac{4}{2 \cdot \sqrt{4 x + 1}}

\frac{2}{\sqrt{4 x + 1}}

▶Derivative:

\frac{2}{\sqrt{4 x + 1}}