यदि y = \(\rm e^{x+e^{x+e^{x+^{\ ...\ \infty}}}}\) तो \(\rm \dfrac{dy}{dx}\) क्या है?

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  1. \(\rm \dfrac{1+y}{y}\)
  2. \(\rm \dfrac{y}{1+y}\)
  3. \(\rm \dfrac{y}{1-y}\)
  4. \(\rm \dfrac{1-y}{y}\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \dfrac{y}{1-y}\)
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Detailed Solution

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संकल्पना:

अवकलजों का शृंखला नियम:

\(\rm \dfrac{d}{dx}f(g(x))=\dfrac{d}{d\ g(x)}f(g(x))\times \dfrac{d}{dx}g(x)\)

\(\rm \dfrac{d}{dx}e^x\) = ex

गणना:

यह दिया गया है कि y = \(\rm e^{x+e^{x+e^{x+^{\ ...\ \infty}}}}\)

∴ y = \(\rm e^{x+(e^{x+e^{x+^{\ ...\ \infty}}})}=e^{x+y}\)

x के संबंध में दोनों पक्षों को अवकलित करके और श्रृंखला नियम का उपयोग करना, हमें मिलता है:

\(\rm \dfrac{dy}{dx}=\dfrac{d}{dx}e^{x+y}\)

\(\rm \dfrac{dy}{dx}=e^{x+y}\dfrac{d}{dx}(x+y)\)

\(\rm \dfrac{dy}{dx}=y\left (1+\dfrac{dy}{dx} \right )\)

\(\rm \dfrac{dy}{dx}=y+y\dfrac{dy}{dx}\)

\(\rm (1-y)\dfrac{dy}{dx}=y\)

\(\rm \dfrac{dy}{dx}=\dfrac{y}{1-y}\)

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