Question

If we have a function as $y = \sqrt {\sin x + y} $ . Then $\dfrac{{dy}}{{dx}} = $
$\left( 1 \right)$ $\dfrac{{\sin x}}{{2y - 1}}$
$\left( 2 \right)$ $\dfrac{{\cos x}}{{2y - 1}}$
$\left( 3 \right)$ $\dfrac{{\sin x}}{{2y + 1}}$
$\left( 4 \right)$ $\dfrac{{\cos x}}{{2y + 1}}$

Answer 1

Hint: We have to find the derivative of $\sqrt {\sin x + y} $ with respect to $x$ . We solve this using chain rule and various basic derivative formulas of trigonometric functions and derivatives of $x$ . We first square the terms and then take the terms of $y$ on one side and the terms on R.H.S. then simply differentiating and simplifying we get the required result .


Complete step-by-step solution:
Chain rule : The chain rule of differentiation states that we derivate a function or an expression in the form of a chain starting with the first one and then moving on respectively . We first differentiate the outermost function of the given expression using the particular rule of the differentiation and then coming to the inner function and so on differentiating the innermost function of the given expression or function . We may have to use the various rules of the derivatives .
Given : $y = \sqrt {\sin x + y} $
Squaring both sides , we get the expression as :
${y^2} = \sin x + y$
Taking the terms of $y$ on one side and $x$ terms on the other , we get the expression as :
${y^2} - y = \sin x$
Now we have to derivative of $y$ with respect to $x$
As we know , ( Derivative of $\sin x = \cos x$)
Also , ( derivative of ${x^n} = n{\left( x \right)^{n - 1}}$)
Using these derivatives
Differentiate $y$ with respect to $x$ , we get the expression as :
$2y \times \dfrac{{dy}}{{dx}} - \dfrac{{dy}}{{dx}} = \cos x$
Taking $\dfrac{{dy}}{{dx}}$ common in L.H.S. , we get
$\left( {2y - 1} \right)\dfrac{{dy}}{{dx}} = \cos x$
$\dfrac{{dy}}{{dx}} = \dfrac{{\cos x}}{{2y - 1}}$
Hence , the derivative of $y$ with respect to $x$ is $\dfrac{{\cos x}}{{2y - 1}}$ .
Thus , the correct option is $\left( 2 \right)$.

Note: We differentiated $y$ with respect to $x$ to find $\dfrac{{dy}}{{dx}}$ . We know the differentiation of trigonometric function :
$\dfrac{d}{{dx}}\cos x = - \sin x$
$\dfrac{d}{{dx}}\sin x = \cos x$
$\dfrac{d}{{dx}}{x^n} = n{\left( x \right)^{n - 1}}$
$\dfrac{d}{{dx}}\tan x = {\sec ^2}x$
We use the derivative according to the given problem.
Here we could also use the chain rule of differentiation where we first differentiate the square root function and after that we differentiate the inner function and then simplify the expression but that would be a long approach and that may lead to the wrong answer.