If y=eax ,show that `xdy/dx=ylogy`
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Solution
`y=e^(ax)`
`y=e^(ax) ...............(i)`
`logy=ax..............(ii)`
`dy/dx=ae^(ax)`
`dy/dx=ay`
`xdy/dx=axy `
`xdy/dx=ylogy `
Concept: Derivatives of Implicit Functions
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