Question

How do you differentiate $\sin x+\cos y=\sin x\cos y$?

Answer 1

Differentiation refers to the rate with which a function changes. So change of that function is found using another parameter with which the change is measured. So we will differentiate y terms with respect to x and solve $\dfrac{dy}{dx}$.
For eg – if we want to differentiate sin x with respect to x, we get,
$\sin x=\dfrac{d}{dx}(\sin x)=\cos x$

Complete step by step answer:
According to the question we have to differentiate $\sin x+\cos y=\sin x\cos y$.
Before starting to differentiate, check what is asked in the question whether it is $\dfrac{dy}{dx}$ or $\dfrac{dx}{dy}$. At first it might hardly seem any different but it is what makes all the difference.
$\dfrac{dy}{dx}$ means that we have to differentiate $y$ with respect to $x$.
Similarly, $\dfrac{dx}{dy}$ means that we have to differentiate $x$ with respect to $y$.
So, in the given question we are asked to differentiate $y$ with respect to $x$, that is, $\dfrac{dy}{dx}$.
Differentiating both sides, we get
$\dfrac{d}{dx}(\sin x+\cos y)=\dfrac{d}{dx}(\sin x\cos y)$
$\dfrac{d}{dx}(\sin x)+\dfrac{d}{dx}(\cos y)=\dfrac{d}{dx}(\sin x\cos y)$
We know that, derivative of $\sin x$ is $\cos x$. And the derivative $\cos y$ will be different, $\cos y$ will be first differentiated and since the variable is y, so the differentiation of y will be $\dfrac{dy}{dx}$.
On the right hand side of the equation, we have $\sin x\cos y$, so it will undergo differentiation using the product rule.
We get,
$\cos x+(-\sin y)\dfrac{dy}{dx}=\sin x\dfrac{d}{dx}(\cos y)+\cos y\dfrac{d}{dx}(\sin x)$
$\cos x-\sin y\dfrac{dy}{dx}=\sin x(-\sin y)\dfrac{dy}{dx}+\cos y(\cos x)$
Now, we will rearrange the equation to get $\dfrac{dy}{dx}$ on one side and the rest on the other side.
$\sin x(\sin y)\dfrac{dy}{dx}-\sin y\dfrac{dy}{dx}=\cos y(\cos x)-\cos x$
$(\sin x(\sin y)-\sin y)\dfrac{dy}{dx}=\cos y(\cos x)-\cos x$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\cos y(\cos x)-\cos x}{(\sin x(\sin y)-\sin y)}=\dfrac{\cos x(\cos y-1)}{\sin y(\sin x-1)}$
Therefore, $\dfrac{dy}{dx}=\dfrac{\cos x(\cos y-1)}{\sin y(\sin x-1)}$

Note:
While solving the question care should be given to the parameter with which the rate of change of the function is asked else it will result in a wrong answer. The differentiation should be carried out step wise else if a step is missed then the entire answer will go wrong.