How do you differentiate #f(t)= sqrt(ln(t)) / (-4t-6)^2# using the quotient rule?

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Answer 1 · 2018-01-11T15:37:39

#f'(t)=((-4t-6)+16tlnt)/(2tsqrt(lnt)(-4t-6)^3)#

#f(t)=uv -> f'(t) = (u'v-uv')/v^2#

So

#u = sqrt(ln(t))-> u'=1/(2tsqrt(ln(t))#

And:

#v=(-4t-6)^2-> v'=-8(-4t-6)#

Using the quotient rule:

#f'(t) = (1/(2tsqrt(lnt))(-4t-6)^2+8sqrt(ln(t))(-4t-6))/(-4t-6)^4#

#=(((-4t-6)^2+16tlnt(-4t-6))/(2tsqrt(lnt)))/(-4t-6)^4#

Simplifying by canceling a factor of #(-4t-6)#:

#=((-4t-6)+16tlnt)/(2tsqrt(lnt)(-4t-6)^3)#