Verified answer

Solution:

Given:

[tex] \tt \longrightarrow y =x + \dfrac{1}{x + \dfrac{1}{x + \dfrac{1}{x} + ...} } [/tex]

Can be written as:

[tex] \tt \longrightarrow y =x + \dfrac{1}{y} [/tex]

Multiplying both sides by y, we get:

[tex] \tt \longrightarrow {y}^{2} =xy + 1[/tex]

[tex] \tt \longrightarrow {y}^{2} - xy - 1 =0[/tex]

Differentiating both sides wrt x, we get:

[tex] \tt \longrightarrow \dfrac{d}{dx} ({y}^{2} - xy - 1) =0[/tex]

[tex] \tt \longrightarrow \dfrac{d}{dx} ({y}^{2}) - \dfrac{d}{dx}(xy) -0 =0[/tex]

[tex] \tt \longrightarrow \dfrac{d}{dx} ({y}^{2}) - \dfrac{d}{dx}(xy) =0[/tex]

[tex] \tt \longrightarrow 2y\dfrac{dy}{dx} - \bigg \{x\dfrac{dy}{dx} + y \dfrac{dx}{dx} \bigg \} =0 \: \: \: \bigg((uv)' =u'v +v'u \bigg)[/tex]

[tex] \tt \longrightarrow 2y\dfrac{dy}{dx} -x\dfrac{dy}{dx} - y =0[/tex]

[tex] \tt \longrightarrow (2y - x)\dfrac{dy}{dx} = y[/tex]

[tex] \tt \longrightarrow \dfrac{dy}{dx} = \dfrac{y}{2y - x} [/tex]

Which is our required answer.

Learn More:

1. Derivative of f(x) using first principle.

[tex]\displaystyle\tt\longrightarrow f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}[/tex]

2. Derivative of standard functions.

[tex]\tt 1.\:\: \dfrac{d}{dx}(k)=0[/tex]

[tex]\tt 2.\:\: \dfrac{d}{dx}(x^n)=nx^{n-1}[/tex]

[tex]\tt 3.\:\: \dfrac{d}{dx}(a^x)=a^x\:ln\:x[/tex]

[tex]\tt 4.\:\: \dfrac{d}{dx}(e^x)=e^x[/tex]

[tex]\tt 5.\:\: \dfrac{d}{dx}(log_{a}x)=\dfrac{1}{x\:ln\:a}[/tex]

[tex]\tt 6.\:\: \dfrac{d}{dx}(ln\:x)=\dfrac{1}{x}[/tex]

[tex]\tt 7.\:\: \dfrac{d}{dx}(|x|)=\dfrac{x}{|x|}[/tex]

3. Derivative of trigonometric functions.

[tex]\tt 1.\:\: \dfrac{d}{dx}(sin\:x)=cos\:x[/tex]

[tex]\tt 2.\:\: \dfrac{d}{dx}(cos\:x)=-sin\:x[/tex]

[tex]\tt 3.\:\: \dfrac{d}{dx}(tan\:x)=sec^2x[/tex]

[tex]\tt 4.\:\: \dfrac{d}{dx}(cot\:x)=-cosec^2x[/tex]

[tex]\tt 5.\:\: \dfrac{d}{dx}(sec\:x)=sec\:x\:tan\:x[/tex]

[tex]\tt 6.\:\: \dfrac{d}{dx}(cosec\:x)=-cosec\:x\:cot\:x[/tex]

4. Derivatives of inverse trigonometric functions.

[tex]\tt 1.\:\: \dfrac{d}{dx}(sin^{-1}x)=\dfrac{1}{\sqrt{1-x^2}}[/tex]

[tex]\tt 2.\:\: \dfrac{d}{dx}(cos^{-1}x)=\dfrac{-1}{\sqrt{1-x^2}}[/tex]

[tex]\tt 3.\:\: \dfrac{d}{dx}(tan^{-1}x)=\dfrac{1}{1+x^2}[/tex]

[tex]\tt 4.\:\: \dfrac{d}{dx}(cot^{-1}x)=\dfrac{-1}{1+x^2}[/tex]

[tex]\tt 5.\:\: \dfrac{d}{dx}(sec^{-1}x)=\dfrac{1}{|x|\sqrt{x^2-1}}[/tex]

[tex]\tt 6.\:\: \dfrac{d}{dx}(cosec^{-1}x)=\dfrac{-1}{|x|\sqrt{x^2-1}}[/tex]

5. Derivatives of hyperbolic trigonometric functions.

[tex]\tt 1.\:\: \dfrac{d}{dx}(sinh\:x)=cosh\:x[/tex]

[tex]\tt 2.\:\: \dfrac{d}{dx}(cosh\:x)=sinh\:x[/tex]

[tex]\tt 3.\:\: \dfrac{d}{dx}(tanh\:x)=sech^2x[/tex]

[tex]\tt 4.\:\: \dfrac{d}{dx}(coth\:x)=-cosech^2x[/tex]

[tex]\tt 5.\:\: \dfrac{d}{dx}(sech\:x)=-sech\:x\:tanh\:x[/tex]

[tex]\tt 6.\:\: \dfrac{d}{dx}(cosech\:x)=-cosech\:x\:coth\:x[/tex]

6. Derivatives of inverse hyperbolic trigonometric functions.

[tex]\tt 1.\:\: \dfrac{d}{dx}(sinh^{-1}x)=\dfrac{1}{\sqrt{x^2+1}}[/tex]

[tex]\tt 2.\:\: \dfrac{d}{dx}(cosh^{-1}x)=\dfrac{1}{\sqrt{x^2-1}}[/tex]

[tex]\tt 3.\:\: \dfrac{d}{dx}(tanh^{-1}x)=\dfrac{1}{1-x^2}[/tex]

[tex]\tt 4.\:\: \dfrac{d}{dx}(coth^{-1}x)=\dfrac{1}{1-x^2}[/tex]

[tex]\tt 5.\:\: \dfrac{d}{dx}(sech^{-1}x)=\dfrac{-1}{x\sqrt{1-x^2}}[/tex]

[tex]\tt 6.\:\: \dfrac{d}{dx}(cosech^{-1}x)=\dfrac{-1}{|x|\sqrt{1+x^2}}[/tex]

7. Fundamental rules of derivative.

[tex]\tt 1.\:\: \dfrac{d}{dx}(cf)=c\dfrac{df}{dx}[/tex]

[tex]\tt 2.\:\: \dfrac{d}{dx}(f\pm g)=\dfrac{df}{dx}\pm\dfrac{dg}{dx}\:\:\:\:\:\:[Sum\: and\:difference\:\:rules][/tex]

[tex]\tt 3.\:\: \dfrac{d}{dx}(fg)=g\dfrac{df}{dx}+f\dfrac{dg}{dx}\:\:\:\:\:\:[Product\:\:Rule][/tex]

[tex]\tt 4.\:\: \dfrac{d}{dx}\bigg(\dfrac{f}{g}\bigg)=\dfrac{f'g-g'f}{g^2}\:\:\:\:\:\:[Quotient\:\:Rule][/tex]

[tex]\tt 5.\:\: \dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\:\:\:\:\:\:[Chain\:\:Rule][/tex]

Answer:

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