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

Question

#f(x) = ((x+5)/(x^2+2))^2#?

Charles Anctil · Accepted Answer

#f'(x)=-(2(x+5)(x^2+10x-2))/(x^2+2)^3# #f(x)# can be written as #f(x)=(x+5)^2/(x^2+2)^2=u/v# The quotient rule says that if #f(x)=g(x)/(h(x))#, then #f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2# In this case: #g(x)=(x+5)^2# #g'(x)=[d/dx(x+5)*d/dx(u^2)],u=(x+5)# #g'(x)=1*2u# #g'(x)=1*2(x+5)# #g'(x)=2(x+5)# #h(x)=(x^2+2)^2# #h'(x)=[d/dx(x^2+2)*d/dx(u^2)],u=(x^2+2)# #h'(x)=2x*2u# #h'(x)=2x*2(x^2+2)# #h'(x)=4x(x^2+2)# #f'(x)=((x^2+2)^(2)2(x+5)-(x+5)^(2)4x(x^2+2))/((x^2+2)^2)^2# #f'(x)=(2(x^2+2)^2(x+5)-4x(x+5)^(2)(x^2+2))/(x^2+2)^4# #f'(x)=(2(x^2+2)(x+5)-4x(x+5)^2)/(x^2+2)^3# #f'(x)=(2(x+5)((x^2+2)-2x(x+5)))/(x^2+2)^3# #f'(x)=(2(x+5)(-x^2-10x+2))/(x^2+2)^3# #f'(x)=-(2(x+5)(x^2+10x-2))/(x^2+2)^3#

Ezra Adams · Answer

undefined To find the derivative of $ f(x) = \left(\frac{x+5}{x^2+2}\right)^2 $, we use the chain rule and power rule.

Let $ u = \frac{x+5}{x^2+2} $.
Then $ f(x) = u^2 $.

Now, differentiate $ u $ with respect to $ x $ to get $ \frac{du}{dx} $.
Next, apply the power rule to $ u^2 $ and multiply by $ \frac{du}{dx} $.

First, find $ \frac{du}{dx} $:
$$ u = \frac{x+5}{x^2+2} $$
$$ \frac{du}{dx} = \frac{(x^2+2)(1) - (x+5)(2x)}{(x^2+2)^2} $$

Now, apply the power rule:
$$ f'(x) = 2u \frac{du}{dx} $$
$$ f'(x) = 2\left(\frac{x+5}{x^2+2}\right) \left(\frac{(x^2+2)(1) - (x+5)(2x)}{(x^2+2)^2}\right) $$

Simplify the expression to get the final derivative $ f'(x) $.