Find y' for y = (sqrt(x-1))/(sqrt((x+2)²) ·sqrt((x+3)³))
Find $y'$ for $y = \dfrac{\sqrt{x-1}}{\sqrt{(x+2)^{2}}\cdot\sqrt{(x+3)^{3}}}$.
Explanation of Method
Fractional exponents made additive by taking logarithms.
Step 1: Rewrite with rational exponents
Strictly $\sqrt{(x+2)^{2}} = |x+2|$; on the natural domain $x \gt 1$ this is simply $x+2$. So
\[ y = \frac{(x-1)^{1/2}}{(x+2)^{1}\,(x+3)^{3/2}}, \qquad x \gt 1. \]
Step 2: Take the natural logarithm
\[ \ln y = \frac12\ln(x-1) - \ln(x+2) - \frac32\ln(x+3). \]
Step 3: Differentiate both sides
Each inner function has derivative $1$, so every term contributes its coefficient over the corresponding linear factor: $\frac{y'}{y} = \frac{1}{2(x-1)} - \frac{1}{x+2} - \frac{3}{2(x+3)}$.
Step 4: Verification
\[x = 2:\quad y = \frac{1}{4\cdot 5^{3/2}} \approx 0.022361\]
\[\text{bracket} = \frac{1}{2} - \frac14 - \frac{3}{10} = -0.05 \;\Longrightarrow\; y' \approx -0.0011180\]
\[\frac{y(2+10^{-5})-y(2-10^{-5})}{2\times 10^{-5}} \approx -0.0011180 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \frac{\sqrt{x-1}}{(x+2)(x+3)^{3/2}}\left[\frac{1}{2(x-1)} - \frac{1}{x+2} - \frac{3}{2(x+3)}\right]\,}\]
Remark
Remark. The exponents $\tfrac12$, $-1$, $-\tfrac32$ reappear verbatim as the numerators in the bracket: for a product of powers of linear factors, $\dfrac{y'}{y}$ is always $\sum \dfrac{\alpha_{k}}{x-a_{k}}$. Writing $\sqrt{(x+2)^{2}}$ as $x+2$ rather than $|x+2|$ would give the wrong sign on $x \lt -2$, but that region is outside the domain here anyway, since $x-1$ must be non-negative.