Question

How do you find the derivative of \[g\left( t \right)=t\sqrt{4-t}\]?

Answer 1

Hint: Consider g (t) as the product of two algebraic functions. Now, apply the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dt}=u\dfrac{dv}{dt}+v\dfrac{du}{dt}\] to find the derivative g’(t). Here, consider, u = t and \[v=\sqrt{4-t}\]. Use the formula: - \[\dfrac{d\left[ \sqrt{f\left( t \right)} \right]}{dt}=\dfrac{1}{2\sqrt{f\left( t \right)}}\times f'\left( t \right)\] to find the derivative of the function assumed as ‘v’. Here f’ (t) is the representation of the derivative of f (t).

Complete step by step solution: Here, we have been provided with the function \[g\left( t \right)=t\sqrt{4-t}\] and we are asked to differentiate it. Here we are going to use the product rule of differentiation to get the answer.
\[\because g\left( t \right)=t\sqrt{4-t}\]
Clearly, we can see that we have g (t) is a function of t. Now, we can assume the given function as the product of two algebraic functions, t and $\sqrt{4-t}$. So, we have,
\[\Rightarrow g\left( t \right)=t\times \sqrt{4-t}\]
Let us assume t as ‘u’ and $\sqrt{4-t}$ as ‘v’. So, we have,
\[\Rightarrow g\left( t \right)=u\times v\]
Differentiating both the sides with respect to t, we get,
\[\Rightarrow \dfrac{d\left[ g\left( t \right) \right]}{dt}=\dfrac{d\left( u\times v \right)}{dt}\]
Now, applying the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dt}=u\dfrac{dv}{dt}+v\dfrac{du}{dt}\], we get,
\[\Rightarrow \dfrac{d\left[ g\left( t \right) \right]}{dt}=\left[ u\dfrac{dv}{dt}+v\dfrac{du}{dt} \right]\]
Substituting the assumed values of u and v, we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ g\left( t \right) \right]}{dt}=\left[ t\dfrac{d\left[ \sqrt{4-t} \right]}{dt}+\sqrt{4-t}\dfrac{dt}{dt} \right] \\
 & \Rightarrow \dfrac{d\left[ g\left( t \right) \right]}{dt}=\left[ t\dfrac{d\left[ \sqrt{4-t} \right]}{dt}+\sqrt{4-t} \right] \\
\end{align}\]
We know that: \[\dfrac{d\left[ \sqrt{f\left( t \right)} \right]}{dt}=\dfrac{1}{2\sqrt{f\left( t \right)}}\times f'\left( t \right)\], so comparing (4 – t) with f (t) we have,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ g\left( t \right) \right]}{dt}=\left[ t\times \dfrac{1}{2\sqrt{4-t}}\times \dfrac{d\left( 4-t \right)}{dt}+\sqrt{4-t} \right] \\
 & \Rightarrow \dfrac{d\left[ g\left( t \right) \right]}{dt}=\left[ \dfrac{t}{2\sqrt{4-t}}\times \left( -1 \right)+\sqrt{4-t} \right] \\
 & \Rightarrow \dfrac{d\left[ g\left( t \right) \right]}{dt}=\left[ \dfrac{-t}{2\sqrt{4-t}}+\sqrt{4-t} \right] \\
\end{align}\]
Hence, the above relation is our answer.

Note: One may note that the function whose derivative we are finding must be defined. You may see that for the given function g (t) to be defined we must have the value of the expression (4 – t) greater than or equal to 0 such that the radical term gets defined. That means t must be less than or equal to 4. Therefore the function is differentiable in the interval $\left( -\infty ,4 \right]$. You must remember all the basic rules and formulas of differentiation like: - the product rule, chain rule, \[\dfrac{u}{v}\] rule etc. as they are frequently used in both differential and integral calculus.