Respuesta :
Answer:
[tex]\frac{dy}{dx}=\frac{2e^{2x}-1}{e^{2x}-1}[/tex]
Step-by-step explanation:
Data provided in the question:
[tex]y =\ln(e^{2x} (e^{2x} - 1)^\frac{1}{2})[/tex]
now,
we know
ln(AB) = ln(A) + ln(B)
Therefore,
[tex]y =\ln(e^{2x}) + \ln((e^{2x} - 1)^\frac{1}{2})[/tex]
also,
ln(aᵇ) = b × ln(a)
thus,
[tex]y =\ln(e^{2x}) + \frac{1}{2}\times\ln((e^{2x} - 1))[/tex]
differentiating with respect to 'x' , we get
[tex]\frac{dy}{dx}=\frac{1}{e^{2x}}\times\frac{d(e^{2x})}{dx}+(\frac{1}{2})(\frac{1}{e^{2x}-1})\times\frac{d(e^{2x}-1)}{dx}[/tex]
[∵ derivative of ln(a)=[tex]\frac{1}{a} \times\frac{d(a)}{dx}[/tex]) ]
[tex]\frac{dy}{dx}=\frac{1}{e^{2x}}\times e^{2x}\times\frac{d(2x)}{dx}+(\frac{1}{2})(\frac{1}{e^{2x}-1})\times(e^{2x}-0)\times\frac{d(2x-1)}{dx}[/tex]
or
[tex]\frac{dy}{dx}=\frac{e^{2x}}{e^{2x}}\times2+(\frac{1}{2})(\frac{1}{e^{2x}-1})\times(e^{2x})\times2[/tex]
or
[tex]\frac{dy}{dx}=1+(\frac{1}{2})(\frac{1}{e^{2x}-1})\times(e^{2x})\times2[/tex]
or
[tex]\frac{dy}{dx}=1+(\frac{1}{2}\times2\times\frac{(e^{2x}}{e^{2x}-1})[/tex]
or
[tex]\frac{dy}{dx}=1+(\frac{(e^{2x}}{e^{2x}-1})[/tex]
or
[tex]\frac{dy}{dx}=\frac{(e^{2x}-1)+e^{2x}}{e^{2x}-1})[/tex]
or
[tex]\frac{dy}{dx}=\frac{2e^{2x}-1}{e^{2x}-1}[/tex]
