Differentiation Question 461
Question: If $ y=\frac{\sqrt{x}{{(2x+3)}^{2}}}{\sqrt{x+1}}, $ then $ \frac{dy}{dx}= $
[AISSE 1986]
Options:
A) $ y[ \frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)} ] $
B) $ y[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{2(x+1)} ] $
C) $ y[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{x+1} ] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=\frac{\sqrt{x}{{(2x+3)}^{2}}}{\sqrt{x+1}}\Rightarrow \log y=\frac{1}{2}\log x+2\log (2x+3)-\frac{1}{2}\log (x+1) $
$ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2x}+\frac{2.2}{(2x+3)}-\frac{1}{2(x+1)} $
or $ \frac{dy}{dx}=y[ \frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)} ] $ .
BETA
