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Question: Answered & Verified by Expert
If $y=(x-1)^{2}(x-2)^{3}(x-3)^{5}$, then $\frac{d y}{d x}$ at $x=4$ is equal to
MathematicsDifferentiationKCETKCET 2021
Options:
  • A 108
  • B 54
  • C 36
  • D 516
Solution:
2635 Upvotes Verified Answer
The correct answer is: 516
$y=(x-1)^{2}(x-2)^{3}(x-3)^{5}$
Taking log on both sides,
$\begin{aligned}
&\Rightarrow \log y=\log \left[(x-1)^{2}(x-2)^{3}(x-3)^{5}\right] \\
&\log y=2 \log (x-1)+3 \log (x-2)+5 \log (x-3)
\end{aligned}$
On both sides differentiating w.r.t. $x$, we get
$\frac{1}{y} \frac{d y}{d x}=\frac{2}{x-1}+\frac{3}{x-2}+\frac{5}{x-3}$
$\Rightarrow \quad \frac{d y}{d x}=(x-1)^{2}(x-2)^{3}(x-3)^{5}$
$\left[\frac{2}{(x-1)}+\frac{3}{(x-2)}+\frac{5}{(x-3)}\right]$
$\therefore\left(\frac{d y}{d x}\right)_{x=4}=3^{2} \times 2^{3} \times 1^{5}\left[\frac{2}{3}+\frac{3}{2}+5\right]$
$=9 \times 8 \times\left(\frac{4+9+30}{6}\right)=516$

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