How do you differentiate #f(x)=x/(x-4)^2# using the quotient rule?

Question

Aurora Alvarado · Accepted Answer

See below.

The quotient rule is given by

In this case, #f(x)=x# and #g(x)=(x-4)^2#. We will also make use of the chain rule.

As indicated, we will first take the derivative of #f(x)# and multiply this by #g(x)#, which we leave as is.

#f'(x)=1#

#g(x)f'(x)=(x-4)^2#

From this, we subtract the product of #g'(x)# and #f(x)#, which we leave as is. We will use the chain rule to differentiate #g(x)#.

#g'(x)=2(x-4)(1)#

#f(x)g'(x)=2x(x-4)#

And, #[g(x)]^2# is #[(x-4)^2]^2=(x-4)^4#

The derivative is then given as:

#f'(x)=((x-4)^2-2x(x-4))/(x-4)^4#

Which simplifies to

#((x-4)-2x)/(x-4)^3#

#=>(-x-4)/(x-4)^3#

#=>f'(x)=-(x+4)/(x-4)^3#

Christopher Agresta · Answer

undefined To differentiate $ f(x) = \frac{x}{(x-4)^2} $ using the quotient rule:

1. Identify $ u(x) = x $ and $ v(x) = (x - 4)^2 $.
2. Apply the quotient rule formula:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

3. Calculate the derivatives of $ u(x) $ and $ v(x) $:
   - $ u'(x) = 1 $ (derivative of $ x $)
   - $ v'(x) = 2(x - 4) $ (derivative of $ (x - 4)^2 $)

4. Plug the derivatives and functions into the quotient rule formula:

$$ f'(x) = \frac{(1)(x - 4)^2 - (x)(2(x - 4))}{(x - 4)^4} $$

5. Simplify the expression:

$$ f'(x) = \frac{(x - 4)^2 - 2x(x - 4)}{(x - 4)^4} $$

6. Expand and combine like terms:

$$ f'(x) = \frac{x^2 - 8x + 16 - 2x^2 + 8x}{(x - 4)^4} $$

$$ f'(x) = \frac{-x^2 + 16}{(x - 4)^4} $$

Thus, the derivative of $ f(x) $ is $ f'(x) = \frac{-x^2 + 16}{(x - 4)^4} $.