If \(y = \displaystyle\frac{(x + 1)^{2}(x + 2)^{3}}{(x + 3)^{4}}\), \(x \gt -1\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(x = 1\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{27}{64}\)
Taking logarithms turns the product and quotient into a sum (logarithmic differentiation):
\(\log_e y = 2\log_e(x + 1) + 3\log_e(x + 2) - 4\log_e(x + 3)\)
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \frac{2}{x + 1} + \frac{3}{x + 2} - \frac{4}{x + 3} \)
At x = 1:
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \frac{2}{2} + \frac{3}{3} - \frac{4}{4} = 1 \)
\( \displaystyle y(1) = \frac{2^{2} \times 3^{3}}{4^{4}} = \frac{108}{256} = \frac{27}{64} \)
So \(\frac{dy}{dx} = y(1) \times 1 = \frac{27}{64}\).