Question

How do you find the derivative of \[s=t\sin t\]?

Answer 1

Hint: Consider ‘s’ in the L.H.S as a function of t and differentiate both the sides with respect to the variable t. Consider ‘s’ as the product of an algebraic function and a trigonometric function. Now, apply the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dt}=u\dfrac{dv}{dt}+v\dfrac{du}{dt}\]. Here, consider, u = t and \[v=\sin t\]. Use the formula: - \[\dfrac{d\sin t}{dt}=\cos t\] to simplify the derivative and get the answer.

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

Note: One may note that whenever we are asked to differentiate a product of two or more functions we apply the product rule. 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. Remember the derivatives of some common functions like: algebraic functions, trigonometric functions, logarithmic functions, exponential functions etc. as we may be asked to find the derivative of the product of any two of these listed functions.