derivative of y=sqrt(2x-1) — step-by-step solution

We save the definition:

y = \sqrt{2 x - 1}

Let's differentiate:

\frac{d}{d x} (\sqrt{2 x - 1})

We differentiate using:

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

(where f =

2 x - 1

)

ⓘ The derivative of sqrt(x) is 1/(2*sqrt(x)):

\frac{d}{d x} (\sqrt{2 x - 1})

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

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

ⓘ Multiplying a fraction by a term multiplies its numerator:

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

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

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

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

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

\frac{d}{d x} (2 x) + 0

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

ⓘ Adding 0 changes nothing:

\frac{d}{d x} (2 x) + 0

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

■A constant factor stays in front of the derivative, (k*u)' = k*u':

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

2

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

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

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

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

▶Derivative:

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