Question

Question #0b1d4

Answer 1

`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))`

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 `(x^2+1)`, it will be subtracted instead of added--an example being:

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

Remembering that the `1/2` will be distributed to each term, this yields:

`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'`, which is the derivative of the original function, multiply both sides of the equation by `y`. Recall that `y` is the original function.

`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))`