Question

Help !!!See question in image.Please show workings .​

Answer 1

Answer:

4x-1

Step-by-step explanation:

The working is shown in the above photo

Answer 2

Answer:

see explanation

Step-by-step explanation:

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

f'(x) = lim ( h tends to zero ) $\frac{f(x+h)-f(x)}{h}$

      = lim ( h to 0 ) $\frac{2(x+h)^2-(x+h)-(2x^2-x)}{h}$

     = lim ( h to zero ) $\frac{2(x^2+2hx+2h^2)-x-h-2x^2+x}{h}$

     = lim ( h to zero ) $\frac{2x^2+4hx+2h^2-h-2x^2}{h}$

     = ( lim h to 0 ) $\frac{4hx+2h^2-h}{h}$

     = ( lim h to 0 ) $\frac{2h(2x+h)-h}{h}$

     = lim ( h to 0 ) $\frac{2h(2x+h)}{h}$ - $\frac{h}{h}$ ← cancel h on numerator/ denominator of both

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

f'(x) = 4x - 1