Find the derivative of $\frac{5-\frac{1}{x^2}}{x+3}$

Question

Test 2 Review #4

Find the derivative of $\frac{5-\frac{1}{x^2}}{x+3}$

$Solution$:

The first step to calculating this derivative is to rewrite it as so:

$\frac{5-x^{-2}}{x+3}$

Now we will apply the quotient rule:

$\frac{d}{dx}(\frac{5-x^{-2}}{x+3})$

$=(\frac{(x+3)(\frac{d}{dx}(5-x^{-2}))-(5-x^{-2})(\frac{d}{dx}(x+3))}{(x+3)^2})$

$=(\frac{(x+3)((\frac{d}{dx}5-\frac{d}{dx}x^{-2}))-(5-x^{-2})((\frac{d}{dx}x+\frac{d}{dx}3))}{(x+3)^2})$

$=(\frac{(x+3)((0-(-2x^{-3}))-(5-x^{-2})((1)+0))}{(x+3)^2})$

$=(\frac{(x+3)(2x^{-3})-(5-x^{-2})}{(x+3)^2})$

$=(\frac{(2x^{-2}+6x^{-3})-5+x^{-2})}{(x+3)^2})$

$=(\frac{2x^{-2}+6x^{-3}-5+x^{-2})}{(x+3)^2})$

$=(\frac{3x^{-2}+6x^{-3}-5}{(x+3)^2})$

This is an acceptable answer. We could multiply the top and bottom by $x^3$:

$=(\frac{3x^{-2}+6x^{-3}-5}{(x+3)^2})(\frac{x^3}{x^3})$

$=(\frac{3x+6-5x^3}{x^3(x+3)^2})$

Answer 1

I like the following way. $$\left(\frac{5-\frac{1}{x^2}}{x+3}\right)'=\frac{\frac{2}{x^3}}{x+3}+\left(5-\frac{1}{x^2}\right)\left(-\frac{1}{(x+3)^2}\right)=$$ $$=\frac{2(x+3)-5x^3+x}{x^3(x+3)^2}=\frac{3x+6-5x^3}{x^3(x+3)^2},$$ which gives the same result.