Question

If $x\sin (a + y) + \sin a\cos (a + y) = 0$ , Prove that $\dfrac{{dy}}{{dx}} = \dfrac{{{{\sin }^2}(a + y)}}{{\sin a}}$.

Answer 1

Hint: This problem is based on differentiation. In this problem initially we need to find the expression for $x$ in terms of $y$. Then on differentiating the expression for $x$ with respect to $y$ we will get expression $\dfrac{{dx}}{{dy}}$, further taking the reciprocal we get $\dfrac{{dy}}{{dx}}$.

Complete step by step answer:
Given, $x\sin (a + y) + \sin a\cos (a + y) = 0$ ……… $\left( 1 \right)$
Required to prove, $\dfrac{{dy}}{{dx}} = \dfrac{{{{\sin }^2}(a + y)}}{{\sin a}}$
On simplifying the above equation that is equation $\left( 1 \right)$ to get $x$
$x = \dfrac{{ - \sin a\cos (a + y)}}{{\sin (a + y)}}$
$x = - \sin a.\cot (a + y)$……….. $\left( 2 \right)$
Differentiating the above equation with respect to $y$ , we get
$\dfrac{{dx}}{{dy}} = \left( { - \sin a} \right)\left( { - \cos e{c^2}(a + y)} \right)$ ……. (Because, $\dfrac{{d(\cot (a + y))}}{{dx}} = - \cos e{c^2}(a + y)$ ).
We know that, $\cos ecx = \dfrac{1}{{\sin x}}$
On further simplification, we get
$\dfrac{{dx}}{{dy}} = \dfrac{{\left( {\sin a} \right)}}{{{{\sin }^2}(a + y)}}$
On taking the reciprocal of above equation,
$\dfrac{{dy}}{{dx}} = \dfrac{{{{\sin }^2}(a + y)}}{{\left( {\sin a} \right)}}$
Hence proved.

Additional information:
$ \bullet $ In calculus, the two important concepts are integration and differentiation. Differentiation is a method of finding the derivative of a function. Differentiation is a process, in Maths, where we find the instantaneous rate of change in function based on its variables. One of the most common examples for differentiation is the rate of change of displacement with respect to time is called velocity.
$ \bullet $ If $y$ is a variable and $x$ is another variable, then the rate of change of $y$ with respect to $x$ is given by $\dfrac{{dx}}{{dy}}$ . This is the general expression of derivative of a function and is represented as ${f^I}(x) = \dfrac{{dx}}{{dy}}$ , where $y = f(x)$ is any function.
$ \bullet$ The three basic derivatives are
i) For algebraic functions, $D({x^n})$ = $n{x^{n - 1}}$ , in which $n$ is any real number.
ii) For trigonometric functions, $D(\sin x) = \cos x$ and $D\left( {\cos x} \right) = - \sin x$ .
iii) For exponential functions, $D\left( {{e^x}} \right) = {e^x}$ .

Note:
It should be noted that the differentiation of constant is zero. The differentiation of constant multiplied or divided with the variable is not zero that is for example, from the equation $\left( 2 \right)$ the differentiation of $x = - \sin a.\cot (a + y)$ is not zero because a constant is multiplied with a variable.