Question #0b1d4
1 Answer
Explanation:
When differentiating logarithmically, first define the function:
y=3sqrt((x(x+1)(x-2))/((x^2+1)(2x+3))y=3√x(x+1)(x−2)(x2+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))))ln(y)=ln(3√x(x+1)(x−2)(x2+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)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))))ln(y)=ln(3)+ln(√x(x+1)(x−2)(x2+1)(2x+3))
Before proceeding, we can also deal with the square root through another property of logarithms:
ln(a^b)=b*ln(a)ln(ab)=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)+ln((x(x+1)(x−2)(x2+1)(2x+3))12)
ln(y)=ln(3)+1/2ln((x(x+1)(x-2))/((x^2+1)(2x+3)))ln(y)=ln(3)+12ln(x(x+1)(x−2)(x2+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)ln(abcd)=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)]ln(y)=ln(3)+12[ln(x)+ln(x+1)+ln(x−2)−ln(x2+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))
