How do you differentiate #f(x)=(1-4x)/(1+4x)# using the quotient rule?

Question

Samantha Adkison · Accepted Answer

#f'(x)=-8/(1+4x)^2#

Given:

#f(x)=(g(x))/(h(x))#

As per the Quotient Rule,

#f'(x)=(g'(x)*h(x)-h'(x)*g(x))/(h^2(x))#

thus:

#f'(x)=((0-4*1)(1+4x)-(1-4x)(0+4))/(1+4x)^2=#

#==(-4-cancel(16x)-4+cancel(16x))/(1+4x)^2=-8/(1+4x)^2#

Ezekiel Alexander · Answer

undefined To differentiate $ f(x) = \frac{{1 - 4x}}{{1 + 4x}} $ using the quotient rule:

1. Identify $ u(x) $ as the numerator and $ v(x) $ as the denominator.
   $ u(x) = 1 - 4x $ and $ v(x) = 1 + 4x $.

2. Apply the quotient rule:
   $$ f'(x) = \frac{{v(x)u'(x) - u(x)v'(x)}}{{[v(x)]^2}} $$

3. Differentiate $ u(x) $ and $ v(x) $ with respect to $ x $:
   $$ u'(x) = -4 $$
   $$ v'(x) = 4 $$

4. Substitute into the quotient rule formula:
   $$ f'(x) = \frac{{(1 + 4x)(-4) - (1 - 4x)(4)}}{{(1 + 4x)^2}} $$

5. Simplify the expression:
   $$ f'(x) = \frac{{-4 - 16x - 4 + 16x}}{{(1 + 4x)^2}} $$
   $$ f'(x) = \frac{{-8}}{{(1 + 4x)^2}} $$

Therefore, the derivative of $ f(x) = \frac{{1 - 4x}}{{1 + 4x}} $ with respect to $ x $ is $ f'(x) = \frac{{-8}}{{(1 + 4x)^2}} $.