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Ответ:

yazanadel56

22.03.2022

Mathematics

$\frac{dy}{dx}=\frac{4x+2}{x^2+x}$

Step-by-step explanation:

Data provided in the question:

$y =\ln(x^{2} (x+1)^\frac{1}{2})$

now,

we know

ln(AB) = ln(A) + ln(B)

Therefore,

$y =\ln(x^{2}) + \ln((x+1)^\frac{1}{2})$

also,

ln(aᵇ) = b × ln(a)

thus,

y = $2\ln(x) + \frac{1}{2}\times\ln(x+1)$

differentiating with respect to 'x' , we get

$\frac{dy}{dx}=2\times\frac{1}{x}+(\frac{1}{2})(\frac{1}{x+1})\times\frac{d(x+1)}{dx}$

[∵ derivative of ln(a) = $\frac{1}{a} \times\frac{d(a)}{dx}$ ) ]

$\frac{dy}{dx}=\frac{2}{x}+(\frac{1}{2})(\frac{1}{x+1})\times1$

$\frac{dy}{dx}=\frac{2}{x}+\frac{1}{2(x+1)}$

$\frac{dy}{dx}=\frac{2\times2(2x+1)}{x\times2(x+1)}$

$\frac{dy}{dx}=\frac{4x+2}{x^2+x}$

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