Answer:
[tex]\frac{dy}{dx}=\frac{4x+2}{x^2+x}[/tex]
Step-by-step explanation:
Data provided in the question:
[tex]y =\ln(x^{2} (x+1)^\frac{1}{2})[/tex]
now,
we know
ln(AB) = ln(A) + ln(B)
Therefore,
[tex]y =\ln(x^{2}) + \ln((x+1)^\frac{1}{2})[/tex]
also,
ln(aᵇ) = b × ln(a)
thus,
y =[tex]2\ln(x) + \frac{1}{2}\times\ln(x+1)[/tex]
differentiating with respect to 'x' , we get
[tex]\frac{dy}{dx}=2\times\frac{1}{x}+(\frac{1}{2})(\frac{1}{x+1})\times\frac{d(x+1)}{dx}[/tex]
[∵ derivative of ln(a) = [tex]\frac{1}{a} \times\frac{d(a)}{dx}[/tex]) ]
or
[tex]\frac{dy}{dx}=\frac{2}{x}+(\frac{1}{2})(\frac{1}{x+1})\times1[/tex]
or
[tex]\frac{dy}{dx}=\frac{2}{x}+\frac{1}{2(x+1)}[/tex]
or
[tex]\frac{dy}{dx}=\frac{2\times2(2x+1)}{x\times2(x+1)}[/tex]
or
[tex]\frac{dy}{dx}=\frac{4x+2}{x^2+x}[/tex]
