Question

Differentiating a Logarithmic Function In Exercise, find the derivative of the function.

y =In(e^2x (e^2x - 1)^1/2)

Answer 1

`(dy)/(dx)=(2e^(2x)-1)/(e^(2x)-1)`

Explanation:

Data provided in the question:

`y =\ln(e^(2x) (e^(2x) - 1)^(1)/(2))`

now,

we know

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

Therefore,

`y =\ln(e^(2x)) + \ln((e^(2x) - 1)^(1)/(2))`

also,

ln(aᵇ) = b × ln(a)

thus,

`y =\ln(e^(2x)) + (1)/(2)*\ln((e^(2x) - 1))`

differentiating with respect to 'x' , we get

`(dy)/(dx)=(1)/(e^(2x))*(d(e^(2x)))/(dx)+((1)/(2))((1)/(e^(2x)-1))*(d(e^(2x)-1))/(dx)`

[∵ derivative of ln(a)=
`(1)/(a) *(d(a))/(dx)`) ]

`(dy)/(dx)=(1)/(e^(2x))* e^(2x)*(d(2x))/(dx)+((1)/(2))((1)/(e^(2x)-1))*(e^(2x)-0)*(d(2x-1))/(dx)`

or

`(dy)/(dx)=(e^(2x))/(e^(2x))*2+((1)/(2))((1)/(e^(2x)-1))*(e^(2x))*2`

or

`(dy)/(dx)=1+((1)/(2))((1)/(e^(2x)-1))*(e^(2x))*2`

or

`(dy)/(dx)=1+((1)/(2)*2*((e^(2x))/(e^(2x)-1))`

or

`(dy)/(dx)=1+(((e^(2x))/(e^(2x)-1))`

or

`(dy)/(dx)=((e^(2x)-1)+e^(2x))/(e^(2x)-1))`

or

`(dy)/(dx)=(2e^(2x)-1)/(e^(2x)-1)`