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

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- Jan 20th 2010, 05:18 PMsurffanDifferentiating
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 - Jan 20th 2010, 06:24 PMarze
for (a) use. when y=uv

$\displaystyle \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$ - Jan 21st 2010, 06:25 AMHallsofIvy
Just use the product rule, as arze says.

Quote:

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

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

**base**be the variable x and the exponent be a constant. Here it is the other way around. The derivative of $\displaystyle a^x$ is $\displaystyle (ln a)a^x$.

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

- Jan 21st 2010, 07:20 AMsurffan
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