What is the domain and derivative of f(x)=ln(2xsqrt(2+x))?
1 answer:
Domain is the numbers yo can use
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0
deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)
so
dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))
domain is all real numbers greater than 0
derivitive is
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0
deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)
so
dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))
domain is all real numbers greater than 0
derivitive is
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Using z-scores, it is found that the value of z is z = 1.96.
-----------------------------
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula, which for a measure X, in a distribution with mean and standard deviation
, is given by:
- It measures how many standard deviations the measure is from the mean.
- Each z-score has an associated p-value, which is the percentile.
- The normal distribution is symmetric, which means that the middle 95% is between the <u>2.5th percentile and the 97.5th percentile</u>.
- The 2.5th percentile is Z with a p-value of 0.025, thus Z = -1.96.
- The 97.5th percentile is Z with a p-value of 0.975, thus Z = 1.96.
- Thus, the value of Z is 1.96.
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