Derivative Help

Iancar29

ONce again im back with some calculus question.


This time im on derivatives.

Heres the function.

f(x)= log ( 3x^2+ 2)^5 * note : should there not be a X beside the log?


Ok im thinking its the product rule.
But the log function is making it intimidating

dy/dx of the right i can do = 5(3x^2+ 2)^4(6x^2)

= 30x^2(3X^2+2)^4

But now what do i do with the log???

Thanks again for those who can help

Fremen

Chain rule. To differentiate f(g(x)), you take the derivative of f(x) and evaluate it at g(x), then multiply that by the derivative of g(x)

y = f(g(x))
dy/dx = f'(g(x)) g'(x)

In your problem, set f(x) equal to log(x). What's the derivative of log(x)?

You made a small mistake differentiating the function inside the log, by the way. Note that you can make your problem easier using

log(a^4) = 4 log(a) before doing anything else.

Edit: actually it depends if you meant

(log ( 3x^2+ 2))^5

or

log (( 3x^2+ 2)^5).

Either way, it's a problem about applying the chain rule.

Iancar29

Fremen wrote: »

Chain rule. To differentiate f(g(x)), you take the derivative of f(x) and evaluate it at g(x), then multiply that by the derivative of g(x)

y = f(g(x))
dy/dx = f'(g(x)) g'(x)

In your problem, set f(x) equal to log(x). What's the derivative of log(x)?

You made a small mistake differentiating the function inside the log, by the way. Note that you can make your problem easier using

log(a^4) = 4 log(a) before doing anything else.

Edit: actually it depends if you meant

(log ( 3x^2+ 2))^5

or

log (( 3x^2+ 2)^5).

Either way, it's a problem about applying the chain rule.

Hey thanks for the reply

BUT theres no log x .... its just log... then the polynomial?
I know that dy/dx of log x = 1/x

Im persuming its (( 3x^2+ 2)^5). as the book doesnt actually give the extra set of brackets.


log(a^4) = 4 log(a) ... wheres this?

Oh crap , stupid mistake indeed.. 5(3x^2+ 2)^4(6x) so tidying that up now i get .

- 30x(3x^2+2)

Still dunno what to with that log.....

Fremen

There's no log(x), but there's log of something more complicated than x - i.e. that polynomial.

So let u = ( 3x^2+ 2)^5 . Now you've got log u.

you want

d/dx log (u)

= (d/du log(u)) * du/dx (this is the chain rule)

and you know how to work out both d/du log(u) and du/dx.

Note that when you work out d/du log(u), your answer comes back in terms of u. but you know that u = ( 3x^2+ 2)^5, so you can sub that back into your expression.

Iancar29

Fremen wrote: »

There's no log(x), but there's log of something more complicated than x - i.e. that polynomial.

So let u = ( 3x^2+ 2)^5 . Now you've got log u.

you want

d/dx log (u)

= (d/du log(u)) * du/dx (this is the chain rule)

and you know how to work out both d/du log(u) and du/dx.

Note that when you work out d/du log(u), your answer comes back in terms of u. but you know that u = ( 3x^2+ 2)^5, so you can sub that back into your expression.

CHeers!! .... I think i can do that now so