How to find the first derivative of f(t)=ln(t)/t^5 ?

Question

campbell_st · Answer

use the quotient rule

$$y =  \frac{f(x)}{g(x)} ....\frac{dy}{dx} = \frac{g(x)f'(x) - f(x)g'(x)}{(f(x))^2}$$

let f(t) = ln(t)     then f'(t) = 1/t

g(t) = t^5                  g'(t) = 5t^4

hope this helps

anonymous · Answer

im having some trouble simplifying

campbell_st · Answer

use the drawing tool to show what you have so far

anonymous · Answer

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campbell_st · Answer

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anonymous · Answer

i got x=0 and x=e^(1/5) for critical numbers

campbell_st · Answer

well it looks like perhaps a different question... 

to the one you has asked...

anonymous · Answer

youre right i forgot to say that i wanted to find the critical numbers with the derivative

campbell_st · Answer

well then the 1st derivative simplifies to 

$$f'(t) = \frac{t^4(1 - 5\ln(t))}{t^{10}}$$

or 

$$f(t) = \frac{ 1 - 5\ln(t)}{t^6}$$

so to find the critical numbers... 

1st you need to realise that t cannot equal 0... as a zero denominator is undefined. 

So a vertical asymptote exists at t = 0

the next critical number will be when 

1 - 5ln(t) = 0

now you need to solve this to find 

so there is only 1 stationary point $$t = e^{\frac{1}{5}}$$

hope this helps

anonymous · Answer

yup the only critical number is at $$t=e^\frac{ 1 }{ 5 }$$