Question

f(x)=sqrt{ln(3x ^{4}}-2x)
help plzz

Answer 1

\[f(x)=\sqrt{\ln(3x ^{4}}-2x)\]

Answer 2

what you need help with ?

Answer 3

derived

Answer 4

well u need to apply chain rule here
when u have a composite function like this

\( f(x)=g(h(x)) \)

then u can write

\( f'(x)= h'(x) \ g'(h(x)) \)

for our particular case \( g(x)=\sqrt{x} \) and \( h(x)=\ln (3x^4-2x) \)

and use this properties

\(\large \frac{d}{dx} \sqrt{u(x)}=\frac{u'(x)}{2\sqrt{u(x)}} \) and \(\large \frac{d}{dx} \ln [u(x)]=\frac{u'(x)}{u(x)} \)

Answer 5

i don't understand:(

Answer 6

my explanation is not well :(

Answer 7

yeah..its difucult

Answer 8

@sami-21

Answer 9

if you take the derivative of
\[ f= \sqrt{u -2x} \]
you get
\[ \frac{df}{dx}= \frac{1}{2}u^{-\frac{1}{2}} \frac{du}{dx} \]

if u happens to be
\[ u= ln(3x^4-2x) \]
can you find du/dx ?

Answer 10

* make the first equation
\[ f= \sqrt{u} \]
not sqrt(u-2x)

Answer 11

Oh Oh... o.O

\( f(x)=\sqrt{\ln (3x^4)}-2x \) ?

Answer 12

not sure, but I am guessing it's ln(3x^4-2x) all under the square root.

Answer 13

Applying the derivatives separately if question given by Kazem is right then..

Answer 14

But I think that whole will be under the root..

Answer 15

ugh I hate my pc. I was drawing it, but it got messed up.

Answer 16

oo