How do you differentiate #g(y) =(2x-5 )e^x # using the product rule?

Answer 1

#(2x-5)e^x+2e^x#

product rule states that: #d/dx(u(x)*v(x))=u'(x)*v(x)+u(x)v'(x)# in your case #u(x)=(2x-5)# and #v(x)=e^x# thereby #u'(x)=2# and #v'(x)=e^x#

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Answer 2

Scarlett Allison

To differentiate ( g(y) = (2x - 5)e^x ) using the product rule, you apply the formula:

[ (uv)' = u'v + uv' ]

Let ( u = 2x - 5 ) and ( v = e^x ). Then differentiate ( u ) with respect to ( x ) to get ( u' ), and differentiate ( v ) with respect to ( x ) to get ( v' ). Finally, plug these values into the product rule formula.

[ u' = 2 ] [ v' = e^x ]

Using the product rule:

[ g'(y) = (2)(e^x) + (2x - 5)(e^x) ] [ g'(y) = 2e^x + (2x - 5)e^x ]

So, ( g'(y) = (2 + 2x - 5)e^x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.