Question

If cos(x + y) = y sin x, then find \[\dfrac{{dy}}{{dx}}\] .

Answer 1

Hint: In this first, we will differentiate on both sides w.r.t x. Differentiating of terms on LHS requires chain rule. So LHS we will use chain rule and then on RHS we will use the product rule to differentiate the terms and finally arrange them to get the required derivative.

Complete step-by-step solution:
In the question, we have
$\cos (x + y) = y \sin x$ and we have to find \[\dfrac{{dy}}{{dx}}\]
Let $f(x) = \cos (x+y)$ . (1)
 Clearly, it is a composite function. To differentiate it, we will use the chain rule.
According to chain rule,
If $F(x) = f(g(x))$ is a composite function then
$F'(x) = f'(g(x).g'(x)$
So, the differentiation of equation can be written as:
f’(x)= $ - \sin (x + y)\left( {1 + \dfrac{{dy}}{{dx}}} \right)$
similarly, for RHS let $g(x) = y \sin x$.
It is in product form.
On applying the product rule of differentiation ,we get:
\[ g’(x)= y \cos x + \sin x \dfrac{{dy}}{{dx}}\]
Now, we have
$\cos (x + y) = y \sin x $
so, putting the value of differentiation in above equation, we get:
$ - \sin (x + y)\left( {1 + \dfrac{{dy}}{{dx}}} \right)$ = \[y \cos x + \sin x \dfrac{{dy}}{{dx}}\]
On rearranging the terms on both sides, we get:
\[\dfrac{{dy}}{{dx}} (-\sin (x+y) – \sin x)\] =$y \cos x +\sin (x+y)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y\cos x + \sin (x + y)}}{{ - (\sin (x + y) + \sin x)}}$

Note: For solving this type of question, you should remember differentiation of basic functions like $\dfrac{{d(\cos x)}}{{dx}} = - \sin x$,$\dfrac{{d(\sin x)}}{{dx}} = \cos x$ ,$\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$etc . You should remember the chain rule and product rule of differentiation. The product rule is given as:
$\dfrac{{d(f(x)g(x))}}{{dx}} = f’(x)g(x)+g’(x)f(x)$