Question

solve this............​

Answer 1

Question :-

$\text{find \:} \sf{ \frac{dy}{dx}} \\ \sf{ y \: = \red{ \sqrt{ \green{\sin \: x + \sin \: x \: + ...... + \infty}} }}$

Answer :-

$\rightarrow \boxed{ \sf{ \red{\frac{dy}{dx}} = \frac{ \green{ \cos \: x}}{ \orange{2y - 1}} }} \:$

Solution :-

According to the question,

$\sf{ \red{ y }\: = \sqrt{ \orange{\sin \: x + \sin \: x \: + ...... + \infty}} } \: \\ \\ \text{ we \: can \: write \: it \: } \\ \\ \rightarrow \sf{ y = \pink{\sqrt{ \blue{\sin \: x }\: + ( \red{ \sin \: x \: + ..... \infty})} }} \\ \\ \rightarrow \sf{ \blue{y} \: = \green{ \sqrt{ \red{ \sin \: x + y}} }} \\ \\ \text{ \orange{squaring }\: on \: \red{both }\: \green{sides} \: } \\ \\ \rightarrow \sf{ \green{{y}^{2} } = \red{ \sin \: x }+ y} \\ \\ \text{now \: \orange{ differentiate} \: \red{with }\: respect \blue{ to \: x}} \\ \\ \rightarrow \sf{ \green{2y \: \frac{dy}{dx} } = \red{\cos \: x} \: + \: \orange{ \frac{dy}{dx} }} \\ \\ \rightarrow \sf{ \orange{2y \: \frac{dy}{dx} }- \frac{dy}{dx} = \green{ \cos \: x}} \\ \\ \rightarrow \sf{ \red{\frac{dy}{dx} } \blue{(2y - 1)} = \pink{\cos \: x}} \\ \\ \rightarrow \boxed{ \sf{ \red{\frac{dy}{dx}} = \frac{ \green{ \cos \: x}}{ \orange{2y - 1}} }}$

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