Find y' for y = 1/(x+ sqrt(x²+3)) - (sqrt(2x+3))/x

Find $y'$ for $y = \dfrac{1}{x+\sqrt{x^{2}+3}} - \dfrac{\sqrt{2x+3}}{x}$.

Step 1: Rationalise the first term

Multiplying numerator and denominator by the conjugate $x-\sqrt{x^{2}+3}$ turns the denominator into a difference of squares: \[\frac{1}{x+\sqrt{x^{2}+3}}\cdot\frac{x-\sqrt{x^{2}+3}}{x-\sqrt{x^{2}+3}} = \frac{x-\sqrt{x^{2}+3}}{x^{2}-\bigl(x^{2}+3\bigr)} = \frac{x-\sqrt{x^{2}+3}}{-3} = \frac{\sqrt{x^{2}+3}-x}{3}.\]

What was a quotient has become a constant multiple of a difference — far easier to differentiate. The function is now

\[ y = \frac{\sqrt{x^{2}+3}-x}{3} - \frac{\sqrt{2x+3}}{x}. \]

Step 2: Differentiate the first term

With $\bigl(\sqrt{x^{2}+3}\bigr)' = \dfrac{x}{\sqrt{x^{2}+3}}$: \[\frac{d}{dx}\!\left(\frac{\sqrt{x^{2}+3}-x}{3}\right) = \frac{1}{3}\!\left(\frac{x}{\sqrt{x^{2}+3}}-1\right) = \frac{x-\sqrt{x^{2}+3}}{3\sqrt{x^{2}+3}}.\]

Step 3: Differentiate the second term

By the quotient rule with $u = \sqrt{2x+3}$, $u' = \dfrac{1}{\sqrt{2x+3}}$, $v = x$, $v' = 1$ — the same cancellation as in Problem 87: \[\frac{d}{dx}\!\left(\frac{\sqrt{2x+3}}{x}\right) = \frac{\dfrac{x}{\sqrt{2x+3}} - \sqrt{2x+3}}{x^{2}} = \frac{x-(2x+3)}{x^{2}\sqrt{2x+3}} = -\frac{x+3}{x^{2}\sqrt{2x+3}}.\]

Step 4: Combine

Subtracting a negative quantity gives a plus sign: $y' = \frac{x-\sqrt{x^{2}+3}}{3\sqrt{x^{2}+3}} + \frac{x+3}{x^{2}\sqrt{2x+3}}$.

Step 5: Verification

\[x = 1:\quad \sqrt{4} = 2,\ \sqrt{5} \approx 2.23607 \;\Longrightarrow\; y' = \frac{1-2}{6} + \frac{4}{2.23607} \approx -0.16667 + 1.78885 = 1.62219\]

\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 1.62219 \;\;\checkmark\]