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

1 Answer
Dec 9, 2016

Answer:

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

Explanation:

To differentiate using the product rule, we must first write the quotient as a product

#y = (2+x)(2-3x)^-1#

I use the product rule in the following order:

#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#

We're finished with the calculus, but we can do some algebra:

# = 1/(2-3x) +(3(2+x))/(2-3x)^2#

# = (2-3x+6+3x)/(2-3x)^2#

# = 8/(2-3x)^2#