Question

How do you find the derivative of \[y=\dfrac{\ln \left( t \right)}{{{t}^{2}}}?\]

Answer 1

Hint: We are given a function as \[y=\dfrac{\ln \left( t \right)}{{{t}^{2}}}.\] We have to differentiate this function. To do so we will observe our function, what type of function it is and then accordingly we will use the differentiation method. We will use the formula \[d\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{{{f}^{'}}\left( x \right)g\left( x \right)-{{g}^{'}}\left( x \right)f\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}.\] We know the derivative of ln(t) is \[\dfrac{1}{t}\] and also \[\dfrac{d\left( {{t}^{n}} \right)}{dt}=n{{t}^{n-1}}.\]

Complete step-by-step solution:
We are given a function as \[y=\dfrac{\ln \left( t \right)}{{{t}^{2}}},\] and we have to differentiate it. We will look closely and we will see that our function comprises two functions say ln (t) and \[{{t}^{2}},\] which are being divided. So where to solve such a problem we will use the quotient rule. The quotient rule says that if we have a function as \[y=\dfrac{u}{v},\] then the derivative of this is given as \[{{y}^{'}}=\dfrac{{{u}^{'}}v-{{v}^{'}}u}{{{v}^{2}}}\] where u’ means the derivative of u and v’ means the derivative of v.
Now, in our function we have \[y=\dfrac{\ln \left( t \right)}{{{t}^{2}}}\] so here we can consider u = ln(t) and \[v={{t}^{2}}.\] Now, applying the quotient rule, we will get,
\[\dfrac{d\left( \dfrac{\ln \left( t \right)}{{{t}^{2}}} \right)}{dt}=\dfrac{{{\left( \ln \left( t \right) \right)}^{‘}}{{t}^{2}}-\ln \left( t \right){{\left( {{t}^{2}} \right)}^{‘}}}{{{\left[ {{t}^{2}} \right]}^{2}}}\]
As the derivative of \[\ln \left( t \right)=\dfrac{1}{t},\] so \[{{\left( \ln \left( t \right) \right)}^{‘}}=\dfrac{1}{t}\] and as the derivative of \[{{t}^{2}}\] is \[\dfrac{d\left( {{t}^{2}} \right)}{dt}=2{{t}^{2-1}}=2t.\] So, using these values above, we will get,
\[\Rightarrow \dfrac{d\left( \dfrac{\ln \left( t \right)}{{{t}^{2}}} \right)}{dt}=\dfrac{\dfrac{1}{t}\times {{t}^{2}}-\ln \left( t \right).2t}{{{t}^{4}}}\]
On simplifying, we get,
\[\Rightarrow \dfrac{d\left( \dfrac{\ln \left( t \right)}{{{t}^{2}}} \right)}{dt}=\dfrac{t-2t\ln \left( t \right)}{{{t}^{4}}}\]
Cancelling t, we get,
\[\Rightarrow \dfrac{d\left( \dfrac{\ln \left( t \right)}{{{t}^{2}}} \right)}{dt}=\dfrac{1-2\ln \left( t \right)}{{{t}^{3}}}\]
So, the derivative of \[y=\dfrac{\ln \left( t \right)}{{{t}^{2}}}\] is \[\dfrac{1-2\ln \left( t \right)}{{{t}^{3}}}.\]

Note: While we look for the derivative of an individual, we need to be clear that we have to state the answer after simplification and fill the very last place. Also, we need to be clear that we will use \[d\left( {{t}^{n}} \right)=n{{t}^{n-1}}\] even if we have n = 1 then \[d\left( t \right)=1{{t}^{1-1}}=1{{t}^{0}}\] as \[{{t}^{0}}=1\] So, we get, d(t) = 1. So, we need to know that the derivative of the constant is always zero.