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Question
Find `dy/dx if y = x^3 – 2x^2 + sqrtx + 1`
Solution
`y = x^3 – 2x^2 + sqrtx +1`
Differentiating w.r.t. x, we get
`dy/dx=d/dx(x^3-2x^2+sqrtx+1)`
=`d/dx(x^3)-2d/dx(x^2)+d/dx(sqrtx)+d/dx(1)`
= `3x^2-2(2x)+d/dx(x^(1/2))+0`
=`3x^2-4x+1/2x^(1/2-1)`
=`3x^2-4x+1/2x^((-1)/2)`
`dy/dx=3x^2-4x+1/(2sqrtx)`
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