Question

Help !!!
See question in image.
Please show workings .
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Answer 1

see explanation

Explanation:

Given f(x) then the derivative f'(x) is

f'(x) = lim ( h tends to zero )
`(f(x+h)-f(x))/(h)`

= lim ( h to 0 )
`(2(x+h)^2-(x+h)-(2x^2-x))/(h)`

= lim ( h to zero )
`(2(x^2+2hx+2h^2)-x-h-2x^2+x)/(h)`

= lim ( h to zero )
`(2x^2+4hx+2h^2-h-2x^2)/(h)`

= ( lim h to 0 )
`(4hx+2h^2-h)/(h)`

= ( lim h to 0 )
`(2h(2x+h)-h)/(h)`

= lim ( h to 0 )
`(2h(2x+h))/(h)` -
`(h)/(h)` ← cancel h on numerator/ denominator of both

= lim( h to 0 ) 2(2x + h) - 1 ← let h go to zero

f'(x) = 4x - 1

Answer 2

4x-1

Explanation:

The working is shown in the above photo