# Question #0b1d4

##### 1 Answer

#### Explanation:

When differentiating logarithmically, first define the function:

#y=3sqrt((x(x+1)(x-2))/((x^2+1)(2x+3))#

Now, take the natural logarithm of both sides.

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

One of logarithm's many abilities is to be able to be split up easily. Here, we can make use of the fact that terms being multiplied inside a logarithm can be split up into two separate logarithms being added, as follows:

#ln(abc)=ln(a)+ln(b)+ln(c)#

This gives us

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

Before proceeding, we can also deal with the square root through another property of logarithms:

#ln(a^b)=b*ln(a)#

Thus, we have

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

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

Now, we can continue splitting up the logarithm. Recall that when there are terms being divided, such as

#ln((ab)/(cd))=ln(a)+ln(b)-ln(c)-ln(d)#

Remembering that the

#ln(y)=ln(3)+1/2[ln(x)+ln(x+1)+ln(x-2)-ln(x^2+1)-ln(2x+3)]#

Now, differentiate both sides of the equation. Recall that differentiation with the natural logarithm function takes the form:

#d/dx(ln(u))=1/u*(du)/dx=(u')/u#

Thus, we obtain

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

Finally, to solve for

#y'=3/2sqrt((x(x+1)(x-2))/((x^2+1)(2x+3)))(1/x+1/(x+1)+1/(x-2)-(2x)/(x^2+1)-2/(2x+3))#