quick calc. question

[destined]

looking for derivative:

function is something like [x^1/2*(x+1)^2(x+1)^3] / (x+1)

i know the product rule is take something like f'(x)g(x)+f(x)g'(x) so should i multiply the sq. root of x to (x+1)^2?

after that i should be fine with the rest of the problem...

if you have any idea what i'm talking about...

 

[SacrosanctFiend]

Do you mean "quiche?" Those are delicious.

 

[Toastedlightly]

just do it the slow way to be sure it is right. Beyond that.. I am lost.

 

[chuckywang]

format your expression with parantheses, please.

 

[Shooters]

Originally posted by: destined
i know the chain rule is take something like f'(x)g(x)+f(x)g'(x)

You're confusing the chain rule and the product rule.

Originally posted by: chuckywang
format your expression with parantheses, please.

That too

 

[destined]

whoopsies. i mean product rule.

 

[DrPizza]

The expression you have above looks like it simplifies first.
Simplify it!!
Then the chain rule is a piece of cake to apply.

isn't it sqrt(x)*(x+1)^4

 

[chuckywang]

Originally posted by: DrPizza
The expression you have above looks like it simplifies first.
Simplify it!!
Then the product rule is a piece of cake to apply.

isn't it sqrt(x)*(x+1)^4

FIXED

 

[Shooters]

If you're not missing any parentheses, then you'll just need to apply the product rule recursively.

 

[DrPizza]

Also, if you have 3 functions,
the derivative is fgh' + fg'h + f'gh

 

[DrPizza]

Originally posted by: chuckywang

Originally posted by: DrPizza
The expression you have above looks like it simplifies first.
Simplify it!!
Then the product rule is a piece of cake to apply.

isn't it sqrt(x)*(x+1)^4

FIXED

Thanks. I stopped thinking for a moment there!

 

[DrPizza]

Originally posted by: DrPizza
Also, if you have 3 functions,
the derivative is fgh' + fg'h + f'gh

--- so, if you're insisting on the chain rule, then when you do the derivative of the numerator (while applying the quotient rule), use what I have above.