Question

ok so im getting this wrong the derivitive of ln 5x = to what? 1/5x? and the derivitve of Sqrt 5x = (5x)^1/2= 1/2(5x)^-1/2?? not sure how or where i went wrong

Answer 1

is that even right?

Answer 2

d/dx ln(5x) = (1/x)

Answer 3

so 5 is gone?

Answer 4

you nearly got the sqrt(5x) correct, you are mising a factor of 5 from the chain rule

Answer 5

ok so how i do that?

Answer 6

\[\frac{d}{dx} \ln(f(x)) = \frac{ f ' (x) }{f(x) }\]

Answer 7

\[\frac{d}{dx} \ln(5x) = \frac{5}{5x} = \frac{1}{x}\]

Answer 8

o ok...what about the sqrt(5x)..

Answer 9

y= (5x)^(1/2)

apply chain rule

bring power down in front, leave the inside of bracket alone , reduce the power by 1 , then multiply the derivative of the inside of the bracket.

Answer 10

Dx(ln(x)) = |Dx/x|

Answer 11

thats what i did to become this:
\[1/2(5x)^{-1/2}\]

Answer 12

no

Answer 13

is that righht?

Answer 14

did u multiply by derivativbe of the inside?

Answer 15

u are missing a factor of 5

Answer 16

i dont think i was taught that yet...

Answer 17

what you mean factors of 5

Answer 18

just means multiply it by 5

Answer 19

Since 5x plays an important part of the equation; it has to be accounted for in the derivative as well; by multiplying it into the derivative thru the "chain rule"

Answer 20

but lim x approaches infinity.....

Answer 21

can you show so i get the idea...

Answer 22

Dx[F(g(x))] = F'(g(x)) * g'(x)

Answer 23

F(x) = sqrt(x)

g(x) = 5x

Dx[sqrt(5x)] = 1/(2sqrt(5x)) * 5 = 5/2sqrt(5x)

Answer 24

alternatively let u= 5x , then \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]

Answer 25

its easier to do it that way for beginners

Answer 26

both explanations have their pros and cons :)

Answer 27

so its 25x?

Answer 28

after a while you can just do it

Answer 29

no

Answer 30

lol wow....i am totally lost

Answer 31

\[\sqrt{5x}\implies\frac{5}{2\sqrt{5x}}\]

Answer 32

I never go back to the u substitutions , I just remeber the chant "bring power down in front, leave the inside of bracket alone , reduce the power by 1 , then multiply the derivative of the inside of the bracket. "

Answer 33

its too long, a waste of time in an exam

Answer 34

its like peeling away the layers of an onion
\[[F(g(h(x)))]' = F'(g(h(x)))*g'(h(x))*h'(x)*x'\]

Answer 35

what the....? never even saw that.....

Answer 36

mmm now i remeber!!