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

Question

Ezra Adams · Accepted Answer

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

About the rule of quotient:

#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 rule of quotients:

#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)#

Evelyn Agard · Answer

undefined To differentiate $ f(t) = \frac{\sqrt{\ln(t)}}{(-4t - 6)^2} $ using the quotient rule, follow these steps:

1. Identify the numerator and denominator functions: $ \sqrt{\ln(t)} $ and $ (-4t - 6)^2 $.
2. Apply the quotient rule, which states that the derivative of the quotient of two functions $ u(t) $ and $ v(t) $ is $ \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} $.
3. Differentiate the numerator and denominator functions individually.
4. Substitute the derivatives and original functions into the quotient rule formula.

Applying this to $ f(t) = \frac{\sqrt{\ln(t)}}{(-4t - 6)^2} $:

- Let $ u(t) = \sqrt{\ln(t)} $, $ v(t) = (-4t - 6)^2 $.
- $ u'(t) = \frac{1}{2\sqrt{\ln(t)}} \cdot \frac{1}{t} $, $ v'(t) = 2(-4t - 6)(-4) $.

Therefore, the derivative of $ f(t) $ using the quotient rule is:

$$ f'(t) = \frac{\frac{1}{2\sqrt{\ln(t)}} \cdot \frac{1}{t} \cdot (-4t - 6)^2 - \sqrt{\ln(t)} \cdot 2(-4t - 6)(-4)}{(-4t - 6)^4} $$