Help finding the second derivative

[tjohn101]

Homework Statement

f(x)= x^(3)e^(x)

Find f'(x) in simplest form.
Find f"(x) in simplest form.

Homework Equations

The Attempt at a Solution

I found the first derivative to be: (using the product rule)
f'(x) = (e^x)[3x^2] + (x^3)[e^x]
f'(x) = 3x^(2)e^(x) + x^(3)e^(x)
f'(x) = x^[2]e^[x](3 + x)

I'm really not sure how to find f"(x) because of the "e". My initial thoughts would be to use the def. of derivatives on the problem, but then I get confused as to whether I should use the unfactored f'(x) or the factored version, which as far as I can tell would require the product rule. If anyone could help me get to f"(x) that would be great. Thank you!

Also, the next part says to find the slope of the line tangent to the graph of f(x) at two different x values. Would I use f'(x) or f"(x) for this?

 

[lanedance]

to find f", just use the product rule twice ie,
(a(bc))' = a'(bc) + a(bc)' = a'(bc) a(b'c + bc')

for the 2nd part, which do you think represents the slope?

 

[rock.freak667]

The Attempt at a Solution

I found the first derivative to be: (using the product rule)
f'(x) = (e^x)[3x^2] + (x^3)[e^x]
f'(x) = 3x^(2)e^(x) + x^(3)e^(x)
f'(x) = x^[2]e^[x](3 + x)

Now I need to find the second derivative. I'm really not sure how to find the derivative of f'(x) because of the "e". If anyone could help me get to f"(x) that would be great. Thank you!

Also, the next part says to find the slope of the line tangent to the graph of f(x) at two different x values. Would I use f'(x) or f"(x) for this?

Write back f'(x) = x^[2]e^[x](3 + x) as e^x(3x²+x³) and use the product law again.

f'(x) or dy/dx gives the gradient function, which is the gradient of the tangent at a point x on the curve f(x).

 

[tjohn101]

Okay, I did what you guys said and ended up with:

f"(x)= x^[2]e^[x](15x^[2] + x + 3)

I don't know if I did it right. What do you guys think?

And now that I think about it it's obvious that f'(x) should be used for the slope.. Thanks guys!

 

[rock.freak667]

I did not get that, post your working showing how you got that.

 

[tjohn101]

I did not get that, post your working showing how you got that.

f'(x)= e^[x](3x^[2] + x^[3])

f"(x)= e^[x](3x^[2] + x^[3]) + e^[x]((6x)(x^[3]) + (3x^[2])(3x^[2]) <---I think I messed up in here

= 3x^[2]e^[x] + x^[3]e^[x] + 15x^[4]e^[x]

=x^[2]e^[x](15x^[2] + x + 3)

I could've messed up anywhere and I'll never see it. Where did I go wrong?

 

[rock.freak667]

f'(x)= e^[x](3x^[2] + x^[3])

f"(x)= e^[x](3x^[2] + x^[3]) + e^[x]((6x)(x^[3]) + (3x^[2])(3x^[2]) <---I think I messed up in here

= 3x^[2]e^[x] + x^[3]e^[x] + 15x^[4]e^[x]

=x^[2]e^[x](15x^[2] + x + 3)

I could've messed up anywhere and I'll never see it. Where did I go wrong?

u=e^x

v=3x²+x³

now just find u(dv/dx)+v(du/dx)

 

[tjohn101]

I ended up with 6xe^[x](x+2)

In case I got it wrong, is the derivative of 3x^[2]+x^[3] 3x^[2]+6x? And how did you type the exponents in there?

 

[rock.freak667]

I ended up with 6xe^[x](x+2)

In case I got it wrong, is the derivative of 3x^[2]+x^[3] 3x^[2]+6x? And how did you type the exponents in there?

use the X² tag in the reply box.

u=e^x du/dx=e^x

v=3x²+x³ dv/dx=6x+3x²

sub those back into the formula. You should get a different answer.

 

[tjohn101]

How does xe^x(x²+6x+6) sound?

 

[rock.freak667]

How does xe^x(x²+6x+6) sound?

Much better.

 

[tjohn101]

Thanks for the help! I caught my simple mistake there at the end.I was using the derivative of v twice in the equation instead of once... : /