# Differentiating

• January 20th 2010, 05:18 PM
surffan
Differentiating
I just want to know if anyone can check to see if i'm solving these questions correctly:

a) y=xe^2x
=no idea

b) f(x)=3^x+x^3
f'(x)=x(3)^x-1 + 3X^2

c) y=e^-(2x+5)
=e^-(2x+5) x ln e x -2

d) y=e^(3x^2 -5x+7)
=e^(3x^2 -5x+7) x ln e x 6x-5
• January 20th 2010, 06:24 PM
arze
for (a) use. when y=uv
$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$
• January 21st 2010, 06:25 AM
HallsofIvy
Quote:

Originally Posted by surffan
I just want to know if anyone can check to see if i'm solving these questions correctly:

a) y=xe^2x
=no idea

Just use the product rule, as arze says.

Quote:

b) f(x)=3^x+x^3
f'(x)=x(3)^x-1 + 3X^2
No. The formula $(x^n)'= n(x^{n-1})$ requires the that base be the variable x and the exponent be a constant. Here it is the other way around. The derivative of $a^x$ is $(ln a)a^x$.

Quote:

c) y=e^-(2x+5)
=e^-(2x+5) x ln e x -2
PLEASE don't use "x" to mean the variable and to mean times!
The derivative is, indeed, e^-(2x+5)(-2)(ln e), but ln e= 1. This is just -2e^-(2x+ 5).

Quote:

d) y=e^(3x^2 -5x+7)
=e^(3x^2 -5x+7) x ln e x 6x-5
Again, don't mix "x" as variable and "times"! Also ln e= 1. This is just (6x-5)e^(3x^2- 5x+ 7).
• January 21st 2010, 07:20 AM
surffan
I gave it a try again and this is what i got

a) y=xe^2x
=e^2x +2xe^2x

b) f(x)= 3^x+x^3
= 3^xln3+3x^2

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

d) y= e^(3x^2-5x+7)
= e^(3x^2-5x+7) (6x-5)

I reviewed the rules and I hope I got it right this time