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

Question

1523 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-09T14:28:09

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

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#