Question

How do you find the derivative of $ y=\sin \left( x+y \right) $ ?

Answer 1

Hint: To find the derivative of the given equation, first of all, we are going to take the derivative with respect to x on both sides then on the R.H.S of the given equation we will use the chain rule to derivate it. Firstly, we will take the derivative of $ \sin \left( x+y \right) $ with respect to x then we multiply the result of this derivative to the derivative of the expression written in the bracket i.e. $ \left( x+y \right) $ .

Complete step by step answer:
The equation given in the above problem is as follows:
 $ y=\sin \left( x+y \right) $
We have to find the derivative of the expression given in the R.H.S of the above equation so we are taking the derivative to both sides of the above equation with respect to x.
 $ \dfrac{dy}{dx}=\dfrac{d\left( \sin \left( x+y \right) \right)}{dx} $ ………… Eq. (1)
We are going to take the derivative of R.H.S of the above equation by using chain rule so first of all we are taking the derivative of $ \sin \left( x+y \right) $ by considering $ \left( x+y \right) $ as $ x $ . So, the derivative of $ \sin \left( x+y \right) $ is equal to $ \cos \left( x+y \right) $ then we will take the derivative of $ \left( x+y \right) $ with respect to x:
Derivative of $ x+y $ with respect to $ x $ is as follows:
 $ \begin{align}
  & \dfrac{d\left( x+y \right)}{dx}=\dfrac{dx}{dx}+\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{d\left( x+y \right)}{dx}=1+\dfrac{dy}{dx} \\
\end{align} $
Now, multiplying the above result with $ \cos \left( x+y \right) $ we will get the derivative of R.H.S of eq. (1).
 $ \dfrac{dy}{dx}=\cos \left( x+y \right)\left( 1+\dfrac{dy}{dx} \right) $
Multiplying $ \cos \left( x+y \right) $ with $ \left( 1+\dfrac{dy}{dx} \right) $ we get,
 $ \dfrac{dy}{dx}=\cos \left( x+y \right)+\cos \left( x+y \right)\dfrac{dy}{dx} $
Rearranging the above equation we get,
 $ \dfrac{dy}{dx}\left( 1-\cos \left( x+y \right) \right)=\cos \left( x+y \right) $
Dividing $ \left( 1-\cos \left( x+y \right) \right) $ on both the sides we get,
 $ \dfrac{dy}{dx}=\dfrac{\cos \left( x+y \right)}{1-\cos \left( x+y \right)} $
Hence, we have got the derivative of the given expression as:
 $ \dfrac{dy}{dx}=\dfrac{\cos \left( x+y \right)}{1-\cos \left( x+y \right)} $

Note:
 You can further simplify the derivative that we have solved above in the following way:
 $ \dfrac{dy}{dx}=\dfrac{\cos \left( x+y \right)}{1-\cos \left( x+y \right)} $
We know that $ 1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2} $ and $ \cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2} $ so using these trigonometric identities in the above equation we get,
 $ \dfrac{dy}{dx}=\dfrac{1-2{{\sin }^{2}}\dfrac{\theta }{2}}{2{{\sin }^{2}}\dfrac{\theta }{2}} $
Simplifying the R.H.S of the above equation we get,
 $ \dfrac{dy}{dx}=\dfrac{1}{2{{\sin }^{2}}\dfrac{\theta }{2}}-1 $
We know that $ \dfrac{1}{{{\sin }^{2}}\dfrac{\theta }{2}}=\text{cose}{{\text{c}}^{2}}\dfrac{\theta }{2} $ so using this relation in the above expression we get,
 $ \dfrac{dy}{dx}=\dfrac{1}{2}\text{cose}{{\text{c}}^{2}}\dfrac{\theta }{2}-1 $