Question

If f(x)= x²+x(2a+2b)+4ab-x. f'(x)=....

DON'T COPY PASTE!!! please use the methods and steps​

Answer 1

From the definition of the derivative,

`f'(x) = \displaystyle \lim_(h\to0) \frac{f(x+h) - f(x)}h`

We have

`f(x) = x^2 + x(2a+2b) + 4ab - x = x^2 + (2a+2b-1)x + 4ab`

and so

`\displaystyle f'(x) = \lim_(h\to0) \frac{\left((x+h)^2 + (2a+2b-1)(x+h) + 4ab\right) - \left(x^2 + (2a+2b-1)x + 4ab\right)}h`

`\displaystyle f'(x) = \lim_(h\to0) \frac{x^2 + 2xh + h^2 + (2a+2b-1)x + (2a+2b-1)h + 4ab - x^2 - (2a+2b-1)x - 4ab}h`

`\displaystyle f'(x) = \lim_(h\to0) \frac{2xh + h^2 + (2a+2b-1)h}h`

`\displaystyle f'(x) = \lim_(h\to0) (2x + h + 2a+2b-1)`

`\displaystyle f'(x) = \boxed{2x + 2a + 2b - 1}`

Answer 2

`f(x) = x^2 +x (2a+2b) +4ab-x\\\\f'(x) = 2x+x \cdot 0 + 2a+2b + 0 -1\\\\~~~~~~~~=2x+0+2a+2b-1\\\\~~~~~~~~=2x+2a+2b-1`