How do you use the product rule to differentiate y= (2+x) /( 2-3x)#?

Question

Aurora Alvarado · Accepted Answer

It is (perhaps) more obvious to use the quotient rule, but we can use the product rule (and the chain rule).

We must first write the quotient as a product in order to use the product rule to differentiate. #y = (2+x)(2-3x)^-1# I apply the following sequence to the product rule: #d/dx(uv) = u'v+uv'#. #dy/dx = overbrace((1))^(u') overbrace((2-3x)^-1)^v + overbrace((2+x))^u overbrace([-1(2-3x)^-2 * d/dx(2-3x)])^(v')# # = (2-3x)^-1 +(2+x)[-(2-3x)^-2(-3)]# # = (2-3x)^-1 +3(2+x)(2-3x)^-2# The calculus is over, but we can still work with algebra: # = 1/(2-3x) +(3(2+x))/(2-3x)^2# # = (2-3x+6+3x)/(2-3x)^2# # = 8/(2-3x)^2#

Cameron Alexander · Answer

undefined To differentiate $ y = \frac{2+x}{2-3x} $ using the product rule, you can follow these steps:

1. Identify the functions $ u $ and $ v $.
   Let $ u = 2 + x $ and $ v = 2 - 3x $.

2. Find the derivatives $ u' $ and $ v' $.
   $ u' = 1 $ and $ v' = -3 $.

3. Apply the product rule formula:
   $ (uv)' = u'v + uv' $.

4. Substitute the values:
   $ (2 + x)(-3) + (2 - 3x)(1) $.

5. Simplify:
   $ -3(2 + x) + (2 - 3x) $.

6. Expand:
   $ -6 - 3x + 2 - 3x $.

7. Combine like terms:
   $ -4 - 6x $.

So, the derivative of $ y $ with respect to $ x $ is $ -4 - 6x $.