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Question
Find `dy/dx if y = "e"^x/logx`
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Solution
y = `"e"^x/logx`
Differentiating w.r.t. x, we get
`dy/dx=d/dx("e"^x/logx)`
= `((logx)d/dx("e"^x) - ("e"^x)d/dx(logx))/(logx)^2`
= `((logx)"e"^x - "e"^x(1/x))/(logx)^2`
= `("e"^x(logx - 1/x))/(logx)^2`
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