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)]#