How do you differentiate #g(y) =(1-x^2)e^(3x) # using the product rule?

Question

1344 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-26T07:25:41

#(dg)/(dx)=e^(3x)(3-2x-3x^2)#

Product rule states that #d/(dx)f(x)g(x)=(df)/(dx)xxg(x)+f(x)xx(dg)/(dx)#

Hence in the given function,

#(dg)/(dx)=(d(1-x^2))/(dx)xxe^(3x)+(1-x^2)xx(d(e^(3x)))/(dx)# or

= #(-2x)xxe^(3x)+(1-x^2)3e^(3x)# or

= #-2xe^(3x)+3e^(3x)(1-x^2)# or

= #e^(3x)(-2x+3(1-x^2))# or

= #e^(3x)(3-2x-3x^2)#