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Question: If \(y = \sin x\) and \(x = 3t\) then \(\dfrac{{dy}}{{dt}}\) will be?...

If y=sin⁡xy = \sin x and x=3tx = 3t then dydt\dfrac{{dy}}{{dt}} will be?

Explanation

Solution

First, in this function y is given and we have to find the derivative of the given function.
The differentiation of the given function is defined as the derivative or rate of change of the function.
The function is said to be differentiable if the limit exists.
Here in this question, we have to find the differentiation of the given function with respect to t.
Formula used:
The chain rule of the differentiation: dydt=dydx×dxdt\dfrac{{dy}}{{dt}} = \dfrac{{dy}}{{dx}} \times \dfrac{{dx}}{{dt}}

Complete step by step answer:
Since given that there are two functions y=sin⁡xy = \sin x and x=3tx = 3t
First, take y=sin⁡xy = \sin x and differentiate the function with respect to x, we get dydx=d(sin⁡x)dx=cos⁡x\dfrac{{dy}}{{dx}} = \dfrac{{d(\sin x)}}{{dx}} = \cos x(while taking the derivative would change the sign of the derivatives in the given limit of the variables, and on the right-hand side the value sin is in the variable of x, hence it is differentiated as with dx) which is the sin rule in differentiation.
For the second value, x=3tx = 3t differentiate with the t, we get dxdt=3\dfrac{{dx}}{{dt}} = 3
Hence from these two values, now applying the terms into the chain rule that dydt=dydx×dxdt\dfrac{{dy}}{{dt}} = \dfrac{{dy}}{{dx}} \times \dfrac{{dx}}{{dt}}
Thus, we get, dydt=dydx×dxdt⇒cos⁡x×3\dfrac{{dy}}{{dt}} = \dfrac{{dy}}{{dx}} \times \dfrac{{dx}}{{dt}} \Rightarrow \cos x \times 3
Hence, we get dydt=3cos⁡x\dfrac{{dy}}{{dt}} = 3\cos x which is the requirement.
Additional information: the inverse process of the differentiation is integration, we will apply the given terms to show that
Let dxdt=3\dfrac{{dx}}{{dt}} = 3, taking integration we get ∫dxdt=∫(3)\int {\dfrac{{dx}}{{dt}}} = \int {\left( 3 \right)} and applying the integration law we get ∫dxdt=∫(3)⇒x=3t\int {\dfrac{{dx}}{{dt}}} = \int {\left( 3 \right)} \Rightarrow x = 3t (if there is only constant in the integration yields the variable t)
Again from the given that, dydx=d(sin⁡x)dx=cos⁡x\dfrac{{dy}}{{dx}} = \dfrac{{d(\sin x)}}{{dx}} = \cos x can be converted into the integration that ∫dydx=∫cos⁡x⇒y=sin⁡x\int {\dfrac{{dy}}{{dx}}} = \int {\cos x} \Rightarrow y = \sin x(the integration of the value cosx is sinx)

Note: The differentiation is the rate of change of the given function at any point, we also applied the chain rule to solve the given function and it is compulsory that if we have two functions and need to convert as the composite of two functions as dydt=dydx×dxdt\dfrac{{dy}}{{dt}} = \dfrac{{dy}}{{dx}} \times \dfrac{{dx}}{{dt}}