a tricky one

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}

8 Answers

1
Che ·

is it this only y = sinx/1 + cosx/1 + sinx/1 + cosx/1 + ............ upto infinty or something else?

1
" ____________ ·

what is y --- whiich series is this

1
aieeee ·

this is the question given. but i'm getting no real approach 2 dis question.

1
Che ·

is 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

1
aieeee ·

no, 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.

1
Che ·

well 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.....

1
" ____________ ·

what is the source yaar

62
Lokesh Verma ·

The 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 :)}

Your Answer

Close [X]