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How do you find the derivative of #g(t)=(10log_4t)/t#?

Answer 1

Luke Alvarez

#(dg)/(dt)=10/ln4*(1-lnt)/t^2#

What is required is to convert base #4# of log to base #e# and then use quotient formula

#g(t)=(10log_4t)/t#

#=(10lnt)/(ln4t)#

Hence #(dg)/(dt)=10/ln4xx(txx1/t-lnt)/t^2#

#=10/ln4*(1-lnt)/t^2#

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Answer 2

Emilia Aguilar

To find the derivative of ( g(t) = \frac{10 \log_4 t}{t} ), we can use the quotient rule of differentiation.

Let ( u = 10 \log_4 t ) and ( v = t ).

Then, using the quotient rule:

[ g'(t) = \frac{v \cdot \frac{d}{dt}(u) - u \cdot \frac{d}{dt}(v)}{v^2} ]

Now, we differentiate ( u ) and ( v ) with respect to ( t ):

[ \frac{d}{dt}(u) = \frac{d}{dt}(10 \log_4 t) = \frac{10}{\ln(4)} \cdot \frac{1}{t \ln(4)} = \frac{10}{t \ln(4)} ]

[ \frac{d}{dt}(v) = \frac{d}{dt}(t) = 1 ]

Now, substitute these derivatives into the quotient rule formula:

[ g'(t) = \frac{t \cdot \frac{10}{t \ln(4)} - 10 \log_4 t \cdot 1}{t^2} ]

[ g'(t) = \frac{10 - 10 \log_4 t}{t^2 \ln(4)} ]

Thus, the derivative of ( g(t) ) is:

[ g'(t) = \frac{10 - 10 \log_4 t}{t^2 \ln(4)} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.