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