Question
Download Solution PDF\(\rm ({1\over 4}, {1\over 4})\) पर वक्र √x + √y = 1 के स्पर्श रेखा का ढलान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
वक्र के स्पर्श रेखा का ढलान:
- f(x) का पहला अवकलज।
- x पर ढलान ज्ञात करने के लिए f '(x) में सांकेतिक बिंदु के x मान को रखने पर।
गणना:
वक्र √x + √y = 1
\(\rm {1\over {2 \sqrt x}} +{1\over 2{\sqrt y}}({dy\over dx}) = 0\)
\(\rm {dy\over dx} = - {{2\sqrt y}\over {2\sqrt x}}\)
\(\rm {dy\over dx}_{({1\over 4}, {1\over 4})} = - 1\)
अतः विकल्प (2) सही है।
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