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Question
Differentiate the following function w.r.t.x. : `2^x/logx`
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Solution
Let y = `2^x/logx`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx" (2^x/logx)`
=` (log x d/dx (2^x) - 2^x d/dx (logx))/(logx)^2`
= `(log x. 2^x . log 2 - 2^x . 1/x)/(log x)^2`
= `[2^x (log x log 2 - 1/x)]/(log x)^2`
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