What is the derivative of #(-x^2+5)/(x^2+5)^2#?

Question

Emilia Aguilar · Accepted Answer

#y' = (-2x(x^2 +5)^2 - 2(-x^2 + 5)(x^2 + 5)(2x))/((x^2 +5)^2)^2# #y' = (-2x(x^2 +5)^2 - 2(-x^2 + 5)(x^2 + 5)(2x))/((x^2 +5)^2)^2# #y' = (-2x(x^4 + 10x +25) - 4x(-x^4 - cancel(5x^2) +cancel(5x^2) + 25))/((x^2 +5)^4# #y' = (-2x^5 - 20x^2 -50x + 4x^5 - 100x)/((x^2 +5)^4# #y' = (2x^5 - 20x^2 - 150x)/((x^2 +5)^4#

Luke Alvarez · Answer

undefined To find the derivative of the function $ \frac{{-x^2 + 5}}{{(x^2 + 5)^2}} $, you can use the quotient rule:

Let $ u = -x^2 + 5 $ and $ v = (x^2 + 5)^2 $.

Then, using the quotient rule, the derivative is:

$$ \frac{{du}}{{dx}} = \frac{{v \cdot \frac{{du}}{{dx}} - u \cdot \frac{{dv}}{{dx}}}}{{v^2}} $$

First, find $ \frac{{du}}{{dx}} $ and $ \frac{{dv}}{{dx}} $:

$$ \frac{{du}}{{dx}} = -2x $$
$$ \frac{{dv}}{{dx}} = 2 \cdot (x^2 + 5) \cdot 2x = 4x \cdot (x^2 + 5) $$

Now, substitute into the quotient rule formula:

$$ \frac{{-x^2 + 5}}{{(x^2 + 5)^2}} = \frac{{(x^2 + 5)^2 \cdot (-2x) - (-x^2 + 5) \cdot 4x \cdot (x^2 + 5)}}{{(x^2 + 5)^4}} $$

Simplify the expression:

$$ = \frac{{-(x^2 + 5)(2x) + (x^2 - 5) \cdot 4x}}{{(x^2 + 5)^3}} $$

$$ = \frac{{-2x^3 - 10x + 4x^3 - 20x}}{{(x^2 + 5)^3}} $$

$$ = \frac{{2x^3 - 10x}}{{(x^2 + 5)^3}} $$

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

So, the derivative of $ \frac{{-x^2 + 5}}{{(x^2 + 5)^2}} $ is $ \frac{{2x(x^2 - 5)}}{{(x^2 + 5)^3}} $.