Question
Download Solution PDFयदि \(\rm x^m y^n =xya^{m+n}\) है, तो \(\rm \frac{dy}{dx}\) किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
माना कि हमारे पास दो फलन f(x) और g(x) हैं और वे दोनों अवकलनीय हैं।
- श्रृंखला नियम:
\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\) - गुणनफल नियम:\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)
- log ab = b log a
गणना:
\(\rm x^m y^n =xya^{m+n}\) , जहाँ a स्थिरांक है।
दोनों पक्षों में log लेने पर, हमें निम्न प्राप्त होता है
⇒ \(\rm \log (x^m y^n) =\log (xya^{m+n})\)
⇒ log xm + log yn = log x + log y + log am+n
⇒ m log x + n log y = log x + log y + (m + n) log a
x के संबंध में दोनों पक्षों का अवकलन करने पर, हमें निम्न प्राप्त होता है
⇒ \(\rm \frac{m}{x}+\frac{n}{y}\frac{\mathrm{d} y}{\mathrm{d} x}= \frac 1 x + \frac 1 y \frac {dy}{dx} + 0\) [∵ \(\rm \frac{d \;\text{constant}}{dx} = 0\) ]
⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}\left [ \frac{n}{y}-\frac{1}{y} \right ]= \left [ \frac{1}{x}- \frac{m}{x} \right ]\)
⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-(m-1)y}{(n-1)x}\) .
सही विकल्प 3 है।
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