Find y' for y = (sqrt(x))/(sqrt(x)+1)

Find $y'$ for $y = \dfrac{\sqrt{x}}{\sqrt{x}+1}$.

Step 1: Identify the parts and apply the quotient rule

For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v-uv'}{v^{2}}$. Since the constant $1$ differentiates to zero, numerator and denominator have the same derivative: \[u = \sqrt{x},\quad u' = \frac{1}{2\sqrt{x}}, \qquad v = \sqrt{x}+1,\quad v' = \frac{1}{2\sqrt{x}}.\]

Step 2: Substitute

\[ y' = \frac{\dfrac{1}{2\sqrt{x}}\bigl(\sqrt{x}+1\bigr) - \sqrt{x}\cdot\dfrac{1}{2\sqrt{x}}}{\bigl(\sqrt{x}+1\bigr)^{2}}. \]

Step 3: Simplify the numerator

Factor out $\dfrac{1}{2\sqrt{x}}$; the two square roots then cancel: \[\frac{1}{2\sqrt{x}}\Bigl[\bigl(\sqrt{x}+1\bigr) - \sqrt{x}\Bigr] = \frac{1}{2\sqrt{x}}\cdot 1 = \frac{1}{2\sqrt{x}}.\]

Step 4: Write the final expression

\[ y' = \frac{\dfrac{1}{2\sqrt{x}}}{\bigl(\sqrt{x}+1\bigr)^{2}} = \frac{1}{2\sqrt{x}\,\bigl(\sqrt{x}+1\bigr)^{2}}. \]

Step 5: Verification

An independent route: write $y = 1 - \dfrac{1}{\sqrt{x}+1}$. Then

\[y' = \frac{\bigl(\sqrt{x}+1\bigr)'}{\bigl(\sqrt{x}+1\bigr)^{2}} = \frac{1}{2\sqrt{x}\,\bigl(\sqrt{x}+1\bigr)^{2}} \;\;\checkmark\]

\[x = 1:\quad y'(1) = \frac{1}{2\cdot 1\cdot 4} = \frac{1}{8} = 0.125; \qquad \frac{y(1.001)-y(0.999)}{0.002} \approx 0.125 \;\;\checkmark\]