Question

Differentiate the following w. r. t. x. : `x sqrtx + logx − e^x`

Answer 1

Let y = `xsqrt x + log x  –  "e"^x`

=`x^(3/2) + log x  –  "e" ^x`

Differentiating w.r.t. x, we get

`dy/dx=d/dx(x^(3/2)+logx - "e"^x)`

= `d/dxx^(3/2)+d/dxlogx-d/dx"e"^x`

= `3/2x^(3/2-1) +1/x - "e"^x`

= `3/2x^(1/2)+1/x - "e"^x`

= `3/2sqrtx+ 1/x - "e"^x`