How do you use the chain rule to differentiate #y=root5(x^2-3)/(-x-5)#?

Question

Emilia Aguilar · Accepted Answer

Separate into components

#y = (root(5)(x^2-3))/(-x-5)# Let #u = root(5)(x^2-3)# and #v =-1/(x+5)# #y' = udv+vdu# #y' = root(5)(x^2-3) imes1/(x+5)^2-1/(x+5)(1/5)(x^2-3)^(-4/5) 2x# #y' = root(5)(x^2-3) imes1/(x+5)^2(1-(2x(x+5))/(x^2-3))# #y' = root(5)(x^2-3) imes1/(x+5)^2(-(x^2+10x+3))/(x^2-3)# #y' = (root(5)(x^2-3))^4 imes(1+22/(x+5)^2)#

Isaiah Allen · Answer

undefined To differentiate $ y = \sqrt{5} \frac{x^2 - 3}{-x - 5} $ using the chain rule, follow these steps:

1. Identify the outer function and the inner function.
2. Apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

The outer function is the square root function, and the inner function is $ \frac{x^2 - 3}{-x - 5} $.

Now, differentiate the outer function with respect to the inner function:
$$ \frac{d}{dx}(\sqrt{5}u) = \sqrt{5} \frac{du}{dx} $$

Where $ u = \frac{x^2 - 3}{-x - 5} $.

Next, differentiate the inner function:
$$ \frac{du}{dx} = \frac{d}{dx} \left( \frac{x^2 - 3}{-x - 5} \right) $$

Use the quotient rule to differentiate:
$$ \frac{du}{dx} = \frac{(2x)(-x - 5) - (x^2 - 3)(-1)}{(-x - 5)^2} $$

Now, substitute $ \frac{du}{dx} $ and $ u $ back into the chain rule equation:
$$ \frac{dy}{dx} = \sqrt{5} \frac{(2x)(-x - 5) - (x^2 - 3)(-1)}{(-x - 5)^2} $$

Simplify the expression if needed. That's your final answer.