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

Answer 1

Elijah Abram

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

Apply the product rule:

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#

#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#

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Answer 2

Emilia Aguilar

To differentiate ( y = (r^2 - 2r)e^r ), we can use the product rule of differentiation.

Let's denote ( u = r^2 - 2r ) and ( v = e^r ).

Then, the derivative of ( u ) with respect to ( r ) is ( \frac{du}{dr} = 2r - 2 ), and the derivative of ( v ) with respect to ( r ) is ( \frac{dv}{dr} = e^r ).

Now, applying the product rule:

[ \frac{dy}{dr} = \frac{du}{dr}v + u\frac{dv}{dr} ]

[ \frac{dy}{dr} = (2r - 2)e^r + (r^2 - 2r)(e^r) ]

[ \frac{dy}{dr} = 2re^r - 2e^r + r^2e^r - 2re^r ]

[ \frac{dy}{dr} = r^2e^r - 2e^r ]

So, the derivative of ( y ) with respect to ( r ) is ( r^2e^r - 2e^r ).

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Answer from HIX Tutor

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