derivatives of f(x)=3xe^x

[kdm]

Can I have some help figuring out the first and second derivatives of f(x) = 3xe^x

 

[galactus]

Use the product rule.

 

[kdm]

f '(x)=3xe^(x)+3e^(x) is this correct for the first derivative?

 

[galactus]

Yep, sure is. Good work. Now, do it again for the second derivative.

 

[kdm]

f "(x)=3e^(x)+(0)e^(x) = 3e^(x)

 

[galactus]

Nope. If you notice, 3xe^x is the original and it's also part of the first derivative.

 

[kdm]

so f "(x)=3xe^x

 

[Guest]

so f "(x)=3xe^x

Actually, I think it would be f"(x)=3xe^x+6e^x because you still have to use the product rule with your answer from the first derivative, which I think gives you
3xe^x+3e^x+3e^x+0 which after combining like terms is 3xe^x+6e^x.

I think?

 

[stapel]

Actually, I think it would be f"(x)=3xe^x+6e^x....

Correct.

Eliz.