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| 1. |
Find dy/dx yx\xa0= xy |
| Answer» Ans. yx\xa0= xyTake log both side, we get\xa0log yx\xa0= log xy=> x log y = y log xdifferentiate w.r.t x, we get=>\xa0\\(x {1\\over y} {dy \\over dx} + log y = y {1\\over x} + log x {dy \\over dx}\\)=>\xa0\\( {x\\over y} {dy \\over dx} - log x {dy \\over dx} = {y\\over x} - log y\\)=>\xa0\\(( {{x\\over y} - log x} ) {dy \\over dx} = {y\\over x} - log y\\)=>\xa0\\( {dy \\over dx} = { ( { {y\\over x} - log y } ) \\over ( {{x\\over y} - log x} )}\\)\xa0 | |