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:

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

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

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

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