How do you differentiate #g(y) =e^x(x^2 + 3) # using the product rule?
1 Answer
Mar 3, 2016
Explanation:
The Product Rule:
#frac{"d"}{"d"x}(uv) = vfrac{"d"u}{"d"x} + ufrac{"d"v}{"d"x}#
In this case,
#u(x) = e^x#
#u'(x) = e^x#
#v(x) = x^2 + 3#
#v'(x) = 2x#
#g(y) = u(x) * v(x)#
#frac{"d"}{"d"x}(g(y)) = v(x) * u'(x) + u(x) * v'(x)#
#= (x^2 + 3) * (e^x) + (e^x) * (2x)#
#= (x^2 + 2x+ 3) * e^x#
However,
#frac{"d"}{"d"y}(g(y)) = frac{"d"}{"d"y}((x^2 + 3) * e^x)#
#= 0#
