Find y' for y = 9/(5(x+2)^5) - 3/((x+2)^4) + 2/((x+2)³)
Find $y'$ for $y = \frac{9}{5(x+2)^{5}} - \frac{3}{(x+2)^{4}} + \frac{2}{(x+2)^{3}} - \frac{1}{2(x+2)^{2}}$.
Explanation of Method
Four negative powers whose derivatives recombine into a perfect cube.
Step 1: Substitute and rewrite as powers
Let $t = x+2$, so that $\dfrac{dt}{dx} = 1$ and the chain rule contributes no extra factor: $y = \tfrac95 t^{-5} - 3t^{-4} + 2t^{-3} - \tfrac12 t^{-2}$.
Step 2: Differentiate term by term
\[ y' = \tfrac95(-5)t^{-6} - 3(-4)t^{-5} + 2(-3)t^{-4} - \tfrac12(-2)t^{-3} = -9t^{-6} + 12t^{-5} - 6t^{-4} + t^{-3}. \]
Step 3: Put over the common denominator $t^{6}$
\[ y' = \frac{-9 + 12t - 6t^{2} + t^{3}}{t^{6}} = \frac{t^{3} - 6t^{2} + 12t - 9}{(x+2)^{6}}. \]
Step 4: Recognise the numerator as a shifted cube
Compare with $(t-2)^{3} = t^{3}-6t^{2}+12t-8$. The numerator is one less: $t^{3}-6t^{2}+12t-9 = (t-2)^{3} - 1$.
Since $t-2 = x$, this is simply
\[ x^{3} - 1 = (x-1)\bigl(x^{2}+x+1\bigr). \]
Step 5: Verification
\[x = 1:\quad y' = \frac{1-1}{3^{6}} = 0, \text{ a critical point} \;\;\checkmark\]
\[x = 0:\quad y' = \frac{-1}{64} = -0.015625; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx -0.015625 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{x^{3}-1}{(x+2)^{6}} = \dfrac{(x-1)\bigl(x^{2}+x+1\bigr)}{(x+2)^{6}}, \qquad x \neq -2\,}\]
Remark
Remark. The coefficients $\tfrac95, -3, 2, -\tfrac12$ were reverse-engineered so that the four derivatives telescope into $x^{3}-1$. Since $x^{2}+x+1 \gt 0$ always and $(x+2)^{6} \gt 0$, the sign of $y'$ is the sign of $x-1$: the function decreases up to $x = 1$ and increases afterwards on the branch $x \gt -2$.