Question

Differentiate with respect to t.
$y=\cos \left( 5-3t \right)$, $\dfrac{dy}{dt}=$
(a) $5\sin \left( 5-3t \right)$
(b) $3\sin \left( 5-3t \right)$
(c) $-5\sin \left( 5-3t \right)$
(d) $-3\sin \left( 5-3t \right)$.

Answer 1

Hint: In order to solve this question, we need to solve this by the chain rule. The chain rule is used to solve in situations where the function is inside a function. The chain rule says that $\dfrac{dy}{dt}=\dfrac{d}{dt}\left( g\left( f\left( t \right) \right) \right)=\dfrac{d\left( g\left( f\left( t \right) \right) \right)}{df\left( t \right)}\times \dfrac{df\left( t \right)}{dt}$ . We can also do this method to as many functions inside a function.

Complete step-by-step answer:
We need to differentiate the expression $y=\cos \left( 5-3t \right)$ with respect to t.
For this calculation, we need to use the chain rule of differentiation.
Let $f\left( t \right)=5-3t$ , and $g\left( f\left( t \right) \right)=\cos \left( 5-3t \right)$
Therefore, $y=g\left( f\left( t \right) \right)$ .
For such types of functions chain rule says that,
$\dfrac{dy}{dt}=\dfrac{d}{dt}\left( g\left( f\left( t \right) \right) \right)=\dfrac{d\left( g\left( f\left( t \right) \right) \right)}{df\left( t \right)}\times \dfrac{df\left( t \right)}{dt}$ .
This rule is called as chain rule.
This is used when we have a function inside a function.
Substituting the values of the function, we get,
$\dfrac{dy}{dt}=\dfrac{d}{dt}\left( \cos \left( 5-3t \right) \right)$
Solving this further we get,
$\dfrac{dy}{dt}=\dfrac{d\left( \cos \left( 5-3t \right) \right)}{d\left( 5-3t \right)}\times \dfrac{d\left( 5-3t \right)}{dt}$
We know that differentiation of $\cos x$ is $-\sin x$.
Therefore, solving this further we get,
$\dfrac{dy}{dt}=\left( -\sin \left( 3-5t \right) \right)\times \dfrac{d\left( 5-3t \right)}{dt}$
Differentiation of the constant is zero and the differentiation of x is one.
Therefore, substituting we get,
$\begin{align}
  & \dfrac{dy}{dt}=\left( -\sin \left( 3-5t \right) \right)\times \left( \dfrac{d\left( 5 \right)}{dt}-3\dfrac{dt}{dt} \right) \\
 & =\left( -\sin \left( 3-5t \right) \right)\times \left( 0-3 \right) \\
\end{align}$
Solving this further we get,
$\begin{align}
  & \dfrac{dy}{dt}=\left( -\sin \left( 3-5t \right) \right)\times \left( -3 \right) \\
 & =3\sin \left( 3-5t \right) \\
\end{align}$
Hence, the correct option is (b).

Note:We cannot do this question without the chain rule, because we have a complicated function inside a cosine function. We can remember this as taking the differentiation of the outermost function without touching the inside function and then taking the differentiation of the inside function without considering the outside function. The chain rule is used to solve much-complicated differentiation. We can see that the constant appearing is 3 after the differentiation because the differentiation of t is 1. So, we can eliminate option (a) and (c).