A) Differentiate from the first principle the f(x) = 6x2 − 10. b) Locate and identify the turning points of the curve y = 4x³−3x²−6x+2. c) Differentiate i. y = 4x³ (3x-1)⁴ ii. 2yx − y² = 25

The derivative $f'(x)$ of a function $f(x)$ is given by:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

First, find $f(x+h)$:

$$f(x+h) = 6(x+h)^2 - 10$$

Expand $(x+h)^2$:

$$f(x+h) = 6(x^2 + 2xh + h^2) - 10$$ $$f(x+h) = 6x^2 + 12xh + 6h^2 - 10$$

Now, find $f(x+h) - f(x)$:

$$f(x+h) - f(x) = (6x^2 + 12xh + 6h^2 - 10) - (6x^2 - 10)$$ $$f(x+h) - f(x) = 6x^2 + 12xh + 6h^2 - 10 - 6x^2 + 10$$ $$f(x+h) - f(x) = 12xh + 6h^2$$

$$\frac{f(x+h) - f(x)}{h} = \frac{12xh + 6h^2}{h}$$

Factor out $h$ from the numerator:

$$\frac{f(x+h) - f(x)}{h} = \frac{h(12x + 6h)}{h}$$

Cancel $h$:

$$\frac{f(x+h) - f(x)}{h} = 12x + 6h$$

$$f'(x) = \lim_{h \to 0} (12x + 6h)$$

Substitute $h=0$:

$$f'(x) = 12x + 6(0)$$ $$f'(x) = 12x$$

The derivative of $f(x) = 6x^2 - 10$ from first principles is $\boxed{12x}$.