1what is y --- whiich series is this
If y = sinx/1 + cosx/1 + sinx/1 + cosx/1 + ............ upto infinty.
then prove ; \frac{dy}{dx} = \frac{(1+y)cosx+sinx}{1+2y+cosx-sinx}
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8 Answers
1is it this only y = sinx/1 + cosx/1 + sinx/1 + cosx/1 + ............ upto infinty or something else?
1this is the question given. but i'm getting no real approach 2 dis question.
1is it wat u meant y=\frac{sinx}{1+\frac{cosx}{1+\frac{sinx}{1+\frac{cosx}{1+.....}}}}
bec this y = sinx/1 + cosx/1 + sinx/1 + cosx/1 + ............ upto infinty doenst mak any sense
1no, dude. question is the same as it looks. no changes.
god knows , if it makes any sense or not.
let the experts try it. they may find some sense out of it.
1well jus try differentiating the exp i gave .....and find dy/dx
it comes out to be same dy/dx that u gave....except der is ysinx in teh numerator instead of jus sinx......(printing mistake must be der)
beleive me ......the expression i gave is the correct form.....
62The expression given in #5 is correct. (#1 is a way to write continued expressions)
The two expressions are equivalent and widely used.
\frac{1}{x+}\frac{1}{x+}\frac{1}{x+}\frac{1}{x+}...\Leftrightarrow \frac{1}{x+\frac{1}{x+\frac{1}{x+\frac{1}{x+...}}}}
\\y=\frac{sin x}{1+\frac{cosx}{1+y}} \\\Rightarrow y=\frac{sin x(1+y)}{1+y+cosx} \\\Rightarrow y+y^2+ycosx=sin x(1+y) \\\text{Differentiate on both sides to get the expression that you want :)}