Find y' for y = sqrt((x+1)³) ·(2x-1)/(x+2) ·
Find $y'$ for $y = \sqrt{(x+1)^{3}}\cdot\dfrac{2x-1}{x+2}\cdot e^{x^{2}}\cdot\sin^{5}x\cos^{2}x$.
Explanation of Method
Six factors turned into six additive terms.
Step 1: Take the natural logarithm
Every product becomes a sum, every quotient a difference, every exponent a coefficient: $\ln y = \frac{3}{2}\ln(x+1) + \ln(2x-1) - \ln(x+2) + x^{2} + 5\ln\sin x + 2\ln\cos x$.
Note that $\ln e^{x^{2}} = x^{2}$ exactly — the exponential simply disappears.
Step 2: Differentiate term by term
Each logarithm contributes $\dfrac{u'}{u}$; the last two give cotangent and tangent as in Problems 119 and 120: \[\frac{y'}{y} = \frac{3}{2(x+1)} + \frac{2}{2x-1} - \frac{1}{x+2} + 2x + \frac{5\cos x}{\sin x} - \frac{2\sin x}{\cos x}.\]
\[ \frac{y'}{y} = \frac{3}{2(x+1)} + \frac{2}{2x-1} - \frac{1}{x+2} + 2x + 5\cot x - 2\tan x. \]
Step 3: Multiply back by $y$
\[ y' = y\left[\frac{3}{2(x+1)} + \frac{2}{2x-1} - \frac{1}{x+2} + 2x + 5\cot x - 2\tan x\right]. \]
Step 4: Verification
Each bracket term can be traced to exactly one factor of the original product: \[\sqrt{(x+1)^{3}} \to \frac{3}{2(x+1)}; \qquad (2x-1) \to \frac{2}{2x-1}; \qquad \frac{1}{x+2} \to -\frac{1}{x+2}\]
\[e^{x^{2}} \to 2x; \qquad \sin^{5}x \to 5\cot x; \qquad \cos^{2}x \to -2\tan x \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \sqrt{(x+1)^{3}}\,\frac{2x-1}{x+2}\,e^{x^{2}}\sin^{5}x\cos^{2}x\left[\frac{3}{2(x+1)} + \frac{2}{2x-1} - \frac{1}{x+2} + 2x + 5\cot x - 2\tan x\right]\,}\]
Remark
Remark. Attempting this with the product and quotient rules directly would require five nested applications; logarithmic differentiation reduces it to one derivative per factor. Strictly, the method computes $\bigl(\ln|y|\bigr)' = \dfrac{y'}{y}$, which is valid wherever $y \neq 0$ regardless of sign — so the answer holds on any interval avoiding the zeros $x = -1$, $x = \tfrac12$, $x = k\pi$ and $x = \tfrac{\pi}{2}+k\pi$, as well as the pole $x = -2$.