Question #f2c8b

2 Answers
Jan 22, 2017

y' = 3/(2(3x + 1)) + 2/(5 - 2x)

Explanation:

First of all, separate the natural logarithms, using the rule ln(a/b) = ln a - ln b.

y = lnsqrt(3x + 1) - ln(5 -2x)

y = ln(3x + 1)^(1/2) - ln(5 - 2x)

Apply the rule lna^n = nlna to get rid of the exponent.

y = 1/2ln(3x + 1) - ln(5 - 2x)

Now differentiate each natural logarithm separately using the chain rule.

For d/dx1/2ln(3x +1)

Let y = 1/2lnu and u = 3x + 1. Then dy/(du) = 1/(2u) and (du)/dx = 3.

Then dy/dx = 1/(2u) * 3 = 3/(2(3x + 1))

For d/dx ln(5 - 2x)

Let y = lnu and u = 5 - 2x. Then dy/(du) = 1/u and (du)/dx = -2.

dy/dx= 1/u * -2

dy/dx = -2/(5 - 2x)

Now subtract them to find the derivative of the entire function.

y' = 3/(2(3x + 1)) - (-2/(5- 2x))

y' = 3/(2(3x + 1)) + 2/(5 - 2x)

Hopefully this helps!

Jan 22, 2017

y' =(6x + 19)/(-12x^2 + 26x + 10)

Explanation:

We are dealing with two function compositions and a fraction, so we will require the use of the chain and quotient rules. The chain rule states that if y is a function of u and u is a function of x, then

dy/(dx) = dy/(du) (du)/dx

The quotient rule states that if a,b two functions of x, then

(d(a/b))/dx = (a'b - ab')/b^2

We know that (lnx)' = 1/x, so if u is a function of x, from the chain rule we have that (lnu)' = 1/u * u':

y' = (5 - 2x)/(sqrt(3x + 1)) * [(sqrt(3x + 1))/(5-2x)]'

Then, to calculate the derivative of the second term, use the quotient rule:

[(sqrt(3x + 1))/(5-2x)]' = [(sqrt(3x + 1))'(5 - 2x) - (sqrt(3x+1))(-2)]/(5-2x)^2

Finally, we will have to find the derivative of sqrt(3x + 1).

Using the chain rule, with v = 3x + 1:

(sqrtv)' = 1/(2sqrtv) * v' = 3/(2sqrt(3x + 1)).

Substituting in the middle expression gives:

[(3(5 - 2x))/(2sqrt(3x + 1)) - (sqrt(3x+1))(-2)]/(5-2x)^2

This looks quite messy. We can simplify by multiplying and dividing the second term in the numerator, by 2sqrt(3x+1), then transform the

fraction into a complex fraction of the form (a/b)/c:

[(3(5 - 2x))/(2sqrt(3x + 1)) - ((3x+1)(-4))/(2sqrt(3x + 1))]/(5-2x)^2

=[[3(5 - 2x) - (3x+1)(-4)]/(2sqrt(3x + 1))]/(5-2x)^2

=[3(5 - 2x) - (3x+1)(-4)]/[(2sqrt(3x + 1))(5-2x)^2]

=(6x + 19)/[2(sqrt(3x + 1))(5-2x)^2]

Much better now. Finally, we can substitute this into the initial expression:

y' = (5 - 2x)/(sqrt(3x + 1)) * (6x + 19)/[2(sqrt(3x + 1))(5-2x)^2]

y' =(6x + 19)/(-12x^2 + 26x + 10)

We could even factor the quadratic on the bottom to get:

y' = (6x + 19)/[(x+1/3)(x-5/2)]