1. ## derivative e^(5x/2)

Can someone please explain to me how or where do I get this answer from doing this derivative equation...

y=(x^2-2x+2)e^(5x/2)

this is where I am stuck:

y=(x^2-2x+2)e^(5x/2)

= (x^2-2x+2)(e^(5x/2))(5/2)+(2x-2)(e^5x/2)

the answer in the book is: (5/2x^2-3x+3)e^(5x/2)

where do you get the 3x and 3 from?

2. ## Re: derivative e^(5x/2)

Originally Posted by nugen
Can someone please explain to me how or where do I get this answer from doing this derivative equation...

y=(x^2-2x+2)e^(5x/2)

this is where I am stuck:

y=(x^2-2x+2)e^(5x/2)

= (x^2-2x+2)(e^(5x/2))(5/2)+(2x-2)(e^5x/2)

the answer in the book is: (5/2x^2-3x+3)e^(5x/2)

where do you get the 3x and 3 from?
$$\large y=\left(x^2-2x+2\right) e^{\dfrac{5x}{2}}$$

using the product rule for derivatives

$$\frac{dy}{dx}=\frac{d}{dx} \left(x^2-2x+2\right) \cdot e^{\dfrac{5x}{2}}+\left(x^2-2x+2\right) \cdot \frac{d}{dx}\left(e^{ \dfrac{5x}{2}}\right) =$$

$$(2x-2)e^{\dfrac{5x}{2}}+\left(x^2-2x+2\right)\frac{5}{2}e^{ \dfrac{5x}{2}} =$$

$$e^{ \dfrac{5x}{2}}\left(\frac{5}{2}\left(x^2-2x+2\right)+(2x-2)\right)=$$

$$e^{ \dfrac{5x}{2}}\left(\frac{5}{2}x^2-3x+3\right)$$

3. ## Re: derivative e^(5x/2)

Multiply everything out. Then combine like terms.

4. ## Re: derivative e^(5x/2)

Your answer is the same as the book's. Pull out the (e^5x/2) term, then simplify this (5/2)(x^2-2x+2) + (2x - 2). You get the same result.
(e^5x/2)[(5/2)(x^2-2x+2) + (2x - 2)] = (e^5x/2)[(5/2)x^2 -5x + 2x + 5 -2]

5. ## Re: derivative e^(5x/2)

Originally Posted by nugen
Can someone please explain to me how or where do I get this answer from doing this derivative equation...
y=(x^2-2x+2)e^(5x/2)
this is where I am stuck:
y=(x^2-2x+2)e^(5x/2)
= (x^2-2x+2)(e^(5x/2))(5/2)+(2x-2)(e^5x/2)
the answer in the book is: (5/2x^2-3x+3)e^(5x/2)
Here is a matter of notation: $e^t=\exp(t)$.

If $y = \left( {{x^2} - 2x + 2} \right)\exp \left( {2.5x} \right)$ then $y' = \left( {2x - 2} \right)\exp \left( {2.5x} \right) + 2.5\left( {{x^2} - 2x + 2} \right)\exp \left( {2.5x} \right)$