y=ln(ln5x)

Question

y=ln(ln5x)

anonymous · Answer

dy/dx??

mathstudent55 · Answer

Remember the chain rule.

anonymous · Answer

i know it's 1/ln5x i just don't know the deriv of   ln 5x    :(

anonymous · Answer

derive of ln5x is (5)/(5x) u prime/u

mathstudent55 · Answer

$ \dfrac{d}{dx} \ln u = \dfrac{1}{u} \dfrac{du}{dx} $

anonymous · Answer

think you can help me with y= 6cos x?

anonymous · Answer

i got -6sinx, but that answer is wrong :( @Spidermonkeyy

anonymous · Answer

y=6cos x is a product rule problem, so F(x) gprime(x) + g(x)fprime(x)
so we have 6 (-sinx) + (cosx x 0) so I would think it is -6sinx

mathstudent55 · Answer

$ \dfrac{d}{dx} [\ln(\ln 5x)] $

$u = \ln 5x$, $\dfrac{du}{dx} = \dfrac{5}{5x} = \dfrac{1}{x} $

$\dfrac{d}{dx} \ln u = \dfrac{1}{u} \dfrac{du}{dx}$

$ \dfrac{d}{dx} [\ln(\ln 5x)] = \dfrac{1}{\ln 5x} \cdot \dfrac{1}{x} = \dfrac{1}{x \ln 5x}$

mathstudent55 · Answer

@Spidermonkeyy 
What rule do you use for $\dfrac{d}{dx} (7x) $ ?

anonymous · Answer

none it's just 7

mathstudent55 · Answer

The rule is:

$\dfrac{d}{dx} (ku) = k \dfrac{du}{dx} $,

where u is a function of x, and k is a constant.

In this case, 

$\dfrac{d}{dx} (7x) = 7 \dfrac{d}{dx} x = 7 \cdot 1 = 7$.

mathstudent55 · Answer

You can do the same with y = 6cos x.

6 is simply a constant, so 

$\dfrac{d}{dx} 6 \cos x = 6 \dfrac{d}{dx} \cos x = 6 \cdot (- \sin x) = -6 \sin x$

You don't need to use the produt rule. Just keeping the 6 as a constant is simpler.
Of course, your answer with the product rule is correct, but you're over complicating the problem.

anonymous · Answer

we didn't learn the theoretic way as I am still in high school so we just learned how to do it but thank you for clearing that up, would you be able to answer my question on related rates?