How do you differentiate #y=(r^2-2r)e^r#?

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Answer 1 · 2017-12-10T03:25:43

#y'=-2e^r+e^rr^2#

The product rule states: #(f*g)'(x)=[f'(x)*g(x)]+[f(x)*g'(x)]#

Let #f(x)=r^2-2r# and #g(x)=e^r# And...

#f'(x)=2r-2#

#g'(x)=e^r#

So

#y'=(2r-2*e^r)+(r^2-2r*e^r)#

Simplify:

#=(2r-2)e^r+(r^2-2r)e^r#

#=cancel(e^r2r)-2e^r+e^rr^2cancel(-e^r2r_#

#=-2e^r+e^rr^2#