Question

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

Answer 1

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,

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