Question

Question #f2c8b

Answer 1

`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!

Answer 2

`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)]`