The simplify an expression as a frection, we need to find eqaul terms in the nominator and denominator, to cancell them. The idea is, if we have a number called n, and a expression like:
[tex]\frac{nx}{ny}[/tex]We can divide n by n in order to get a simplifyed expression.
In this case we have:
[tex]\frac{d}{dx}(\frac{-4x^2+16}{(x^2+4)^2})=\frac{(x^2+4)^2\cdot(-8x)-4x(x^2+4)(-4x+16)}{(x^2+4)^4}[/tex]We can see that the term (x² + 4) is repeated in the numerator and denominator. To do this easier, let's separate this in a sum of fractions:
[tex]\frac{(x^2+4)^2\cdot(-8x)-4x(x^2+4)(-4x+16)}{(x^2+4)^4}=\frac{(x^2+4)^2\cdot(-8x)}{(x^2+4)^4}-\frac{4x(x^2+4)(-4x+16)}{(x^2+4)^4}[/tex]Now it's much easier to cancell the repeated terms:
In the first fraction, we have the parentheses squared in the numerator and power of 4 in the denominator, to divide it, we can use the power propierties:
[tex]\frac{a^n}{a^m}=a^{n-m}[/tex]Then:
[tex]\frac{(x^2+4)^2}{(x^2+4)^4}=(x^2+4)^{2-4}=(x^2+4)^{-2}=\frac{1}{(x^2+4)^2}[/tex]And now the first fraction is:
[tex]\frac{(x^2+4)^2\cdot(-8x)}{(x^2+4)^4}=\frac{1}{(x^2+4)^2}\cdot(-8x)=-\frac{8x}{(x^2+4)^2}[/tex]For the second fraction is very similar:
[tex]\frac{4x(x^2+4)(-4x+16)}{(x^2+4)^4}[/tex]We have the same term (x² + 4) , in the numerator and denominator. Then divide:
[tex]\frac{(x^2+4)}{(x^2+4)^4}=(x^2+4)^{1-4}=(x^2+4)^{-3}=\frac{1}{(x^2+4)^3}[/tex]Then the second fraction is:
[tex]4x(-4x+16)\cdot\frac{1}{(x^2+4)^3}=\frac{4x(-4x^2+16)}{(x^2+4)^3}[/tex]Now we can add the two fraction to get the final asnwer:
[tex]\frac{d}{dx}(\frac{-4x^2+16}{(x^2+4)^2})=-\frac{8x}{(x^2+4)^2}-\frac{4x(-4x^2+16)}{(x^2+4)^3}[/tex]And that's all the simplifying we can do with this derivative
If we look at the second fraction, we have a difference of squares:
[tex](16-4x^2)=(4^2-2x^2)[/tex]Thus:
[tex](4^2-2x^2)=(4-2x)(4+2x)[/tex]