Answer 1

Let y = `2^x/logx`

Differentiating w.r.t. x, we get

`dy/dx=d/dx(2^x/logx)`

=` (logxd/dx(2^x) - 2^xd/dx(logx))/(logx)^2`

= `(logx(2^xlog2) - 2^x(1/x))/(logx)^2`

= `((2^xlogx*log2)(-1/x))/(logx)^2`