Question
Download Solution PDFजर \(\rm x^m y^n =a^{m+n}\), तर मग \(\rm \dfrac{dy}{dx}\) बरोबर किती?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
समजा आपल्याकडे f(x) आणि g(x) अशी दोन फंक्शन आहेत आणि टी दोन्ही भिन्न आहेत.
- चेन नियम: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
- गुणाकार नियम: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)
गणना:
दिले आहे: \(\rm x^m y^n =a^{m+n}\)
x च्या संदर्भात फरक केल्यास, आपल्याला मिळते
\(\Rightarrow \rm x^m\dfrac{dy^n}{dx}+ y^n\dfrac{dx^m}{dx} = \dfrac{da^{m+n}}{dx}\\\Rightarrow \rm x^m \times ny^{n-1} \dfrac{dy}{dx}+ y^n \times mx^{m-1} = 0\\\Rightarrow \rm x^m \times ny^{n-1} \dfrac{dy}{dx}=- y^n \times mx^{m-1} \\ \Rightarrow\dfrac{dy}{dx}=\frac{- y^n \times mx^{m-1}}{\rm x^m \times ny^{n-1}}\\\therefore \dfrac{dy}{dx}= \frac{-my}{nx} \)
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