1. ## Differentiate

hello everyone,

I have to differentiate this using the compsite rule only:
f(x)=e^x(4-x)/6

I am using k'(x)=g'(f(x)) f'(x)

where k(x)=g(f(x)) and

u = f(x) = x(4-x)/6 ,which I multiplied out to give 2/3x - 1/6x^2

so, f'(x) =2/3 - 2/3x

g(u)=e^u

so, k'(x) = (e^u) (2/3 - 2/3x)

stuck from there, also not sure if all of the above is right. Help please.

sweeties

2. Originally Posted by Sweeties
hello everyone,

I have to differentiate this using the compsite rule only:
f(x)=e^x(4-x)/6

I am using k'(x)=g'(f(x)) f'(x)

where k(x)=g(f(x)) and

u = f(x) = x(4-x)/6 ,which I multiplied out to give 2/3x - 1/6x^2

so, f'(x) =2/3 - 1/3x << SiMoon says : it's 2/6 x=1/3, not 2/3

g(u)=e^u

so, k'(x) = (e^u) (2/3 - 2/3x)

stuck from there, also not sure if all of the above is right. Help please.

sweeties
k'(x) = (e^u) (2/3 - 1/3x)

And u=x(4-x)/6, remember

3. Originally Posted by Moo
k'(x) = (e^u) (2/3 - 1/3x)

And u=x(4-x)/6, remember
Yeh, i know, but i couldn't multiply it out
not sure how to do it.
i had a go and got something horrid like:

2/3 e ^x(4-x)/6 - 1/3 ex ^ x(4-x)/6

4. Originally Posted by Sweeties
Yeh, i know, but i couldn't multiply it out
not sure how to do it.
i had a go and got something horrid like:

2/3 e ^x(4-x)/6 - 1/3 ex ^ x(4-x)/6
The red thing means that it's e*x^(...), which is false. It's x*e^(...)

Plus, use parenthesis... You should write e^(x(4-x)/6), so that it's correct.

Keep it in the form $\displaystyle (\frac 23-\frac 13 x)e^{x(4-x)/6}$

It's not so horrible this way

5. oooh, you're a darling moo!
i think i have just been studying too long today, need some sunshine!

6. Originally Posted by Sweeties
oooh, you're a darling moo!
i think i have just been studying too long today, need some sunshine!
Never study too long at once, get some fresh air :

enjoy

7. i forgot the second part:

use the product rule to differentiate the same original function and use previous answer to show that g(x) =xe^(x(4-x)/6) has the derivative
g'(x)=1/3 (3+2x-x^2)e^(x(x-4)/6)

i really dont know how to do this, i tried but got nowhere.

8. Originally Posted by Sweeties
i forgot the second part:

use the product rule to differentiate the same original function and use previous answer to show that g(x) =xe^(x(4-x)/6) has the derivative
g'(x)=1/3 (3+2x-x^2)e^(x(x-4)/6)

i really dont know how to do this, i tried but got nowhere.
I'm not sure... Are you asked to differentiate g by using the product rule ?

Look at this :

$\displaystyle g(x)=x \cdot f(x)$, right ?

What does the product rule says ?
$\displaystyle (uv)'=u'v+uv'$

Where is the product here ?

9. it is asking to show that g(x) =xe^(x(4-x)/6) has the derivative
g'(x)=1/3 (3+2x-x^2)e^(x(x-4)/6) by using the previous answer and by using the product rule where the product rule is:

k'(x) = f'(x) g(x) + f(x) g'(x)

10. Originally Posted by Sweeties
it is asking to show that g(x) =xe^(x(4-x)/6) has the derivative
g'(x)=1/3 (3+2x-x^2)e^(x(x-4)/6) by using the previous answer and by using the product rule where the product rule is:

k'(x) = f'(x) g(x) + f(x) g'(x)
Ok then, we know that g(x)=x*f(x), because f(x)=e^(x(4-x)/6)

--> g'(x)=(x)'*f(x)+x*f '(x)=f(x)+x*f '(x)

Do you agree ?

11. ok. get this part:

g(x) = x.f(x) = xe^(x(4-x)/6)

so

g'(x) = x'.f(x) + x.f'(x)

giving

e^(x(4-x)/6) + x . e^(x(4-x)/6) (2/3 - 1/3x) =

e^(x(4-x)/6) + x . e^(x(4-x)/6) . 1/3(2-x)

but I don't see how this will get to
=1/3 (3+2x-x^2)e^(x(x-4)/6)

12. Originally Posted by Sweeties
e^(x(4-x)/6) + x . e^(x(4-x)/6) . 1/3(2-x)
Yes !

Now, observe :

e^(x(4-x)/6) + x . e^(x(4-x)/6) . 1/3(2-x)

=e^(x(4-x)/6) (1+x(1/3(2-x)))

=e^(x(4-x)/6) * (1+2/3 x - x²)

=...

13. Originally Posted by Moo
Yes !

Now, observe :

e^(x(4-x)/6) + x . e^(x(4-x)/6) . 1/3(2-x)

=e^(x(4-x)/6) (1+x(1/3(2-x)))
sorry, i cant see how you went from the first to the second line

14. Originally Posted by Sweeties
sorry, i cant see how you went from the first to the second line
I factorised

15. aha! so you did.