How do you differentiate #y=(e^x(x^2-7))#?

Question

Paisley Johnson · Accepted Answer

#f'(x)=e^x(x^2+2x-7)# You have to use the Product Rule #f'(x)=uv'+u'v# #u=e^x# #u'=e^x# #v=x^2-7# #v'=2x# #f'(x)=e^x2x+e^x(x^2-7)# Factor out #e^x# #f'(x)=e^x(x^2+2x-7)#

Cameron Alexander · Answer

undefined To differentiate $ y = (e^x)(x^2 - 7) $, you can use the product rule of differentiation.

The product rule states that if you have two functions $ u(x) $ and $ v(x) $, the derivative of their product is given by:

$$ \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x) $$

Applying the product rule to $ y = (e^x)(x^2 - 7) $, where $ u(x) = e^x $ and $ v(x) = x^2 - 7 $:

$$ y' = (e^x)(x^2 - 7)' + e^x(x^2 - 7)' $$

Now, differentiate each term separately:

1. $ (x^2 - 7)' = 2x $
2. $ (e^x)' = e^x $

So,

$$ y' = (e^x)(2x) + e^x(2x) $$
$$ y' = 2xe^x + 2xe^x $$
$$ y' = 4xe^x $$

Therefore, the derivative of $ y = (e^x)(x^2 - 7) $ is $ y' = 4xe^x $.

Adam Adkins · Answer

undefined To differentiate $ y = e^x(x^2 - 7) $, you would use the product rule of differentiation.

First, differentiate $ e^x $, which is simply $ e^x $. Then, apply the product rule to differentiate $ x^2 - 7 $. The derivative of $ x^2 - 7 $ is $ 2x $.

So, the derivative of $ y $ with respect to $ x $ is:

$$ \frac{dy}{dx} = e^x(x^2 - 7) + e^x(2x) $$

$$ = e^x(x^2 - 7 + 2x) $$

$$ = e^x(x^2 + 2x - 7) $$