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Differentiation

Section: Mathematics

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Question : 4 of 34

Marks: +1, -0

If

y=\sqrt{\frac{x^4 \sqrt{3x-5}}{(x^2-3)(2x-3)}}

\left(\frac{dy}{dx}\right)_{x=2}=

[AP EAPCET 24 May 2025 S1]

5
0
1
$-5$

Solution:

y=\sqrt{\frac{x^4 \sqrt{3x-5}}{(x^2-3)(2x-3)}}

Take the natural

\log

to both sides, we get

\ln(y)=\frac{1}{2}\ln\left(\frac{x^4\sqrt{3x-5}}{(x^2-3)(2x-3)}\right)

\ln(y)=\frac{1}{2}\left[\ln(x^4)+\ln(\sqrt{3x-5})-\ln(x^2-3)-\ln(2x-3)\right]

=\frac{1}{2}\left[4\ln(x)+\frac{1}{2}\ln(3x-5)-\ln(x^2-3)-\ln(2x-3)\right]

Now, differentiating w.r.t

x

, we get

\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{2} \left[ \frac{4}{x} + \frac{1}{2} \cdot \frac{3}{3x-5} - \frac{2x}{x^2-3} - \frac{2}{2x-3} \right]

\frac{dy}{dx} = \frac{y}{2} \left[ \frac{4}{x} + \frac{3}{2(3x-5)} - \frac{2x}{x^2-3} - \frac{2}{2x-3} \right]

Now, at

x=2

y=\sqrt{\frac{2^4 \sqrt{3\cdot 2-5}}{(2^2-3)(4-3)}}=\sqrt{\frac{16 \cdot 1}{1 \cdot 1}}=4

So,

\left.\frac{dy}{dx}\right|_{x=2}=\frac{4}{2}\left[ \frac{4}{2}+\frac{3}{2(3\cdot 2-5)}-\frac{2\cdot2}{2^2-3}-\frac{2}{2\cdot 2-3} \right]

=2\left[2+\frac{3}{2}-\frac{4}{1}-2\right]=2\left[\frac{3-8}{2}\right]=-5

\therefore \left.\frac{dy}{dx}\right|_{x=2}=-5

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