Question

find from the first principle the derivative with respect to x of the function y equals to x square minus x + 3​

Answer 1

To find the derivative of the function y = x^2 - x + 3 using the first principle, we start by applying the definition of the derivative:

f'(x) = lim (h -> 0) [f(x+h) - f(x)] / h

where f(x) = x^2 - x + 3.

Now we substitute the function into the equation and simplify:

f'(x) = lim (h -> 0) [(x+h)^2 - (x+h) + 3 - (x^2 - x + 3)] / h

f'(x) = lim (h -> 0) [(x^2 + 2xh + h^2 - x - h + 3) - (x^2 - x + 3)] / h

f'(x) = lim (h -> 0) [2xh + h^2 - h] / h

Now we can cancel out the h in the numerator and denominator, leaving:

f'(x) = lim (h -> 0) [2x + h - 1]

Finally, we take the limit as h approaches 0:

f'(x) = 2x - 1

Therefore, the derivative of y = x^2 - x + 3 with respect to x is f'(x) = 2x - 1.

Explanation:

Answer 2

`\frac{\text{d}y}{\text{d}x}=2x-1`

Explanation:

Differentiating from First Principles is a technique to find an algebraic expression for the gradient at a particular point on the curve.

`\boxed{\begin{minipage}{5.6 cm}\underline{Differentiating from First Principles}\\\\\\$\text{f}\:'(x)=\displaystyle \lim_(h \to 0) \left[\frac{\text{f}(x+h)-\text{f}(x)}{(x+h)-x}\right]$\\\\\end{minipage}}`

The point (x + h, f(x + h)) is a small distance along the curve from (x, f(x)). As h gets smaller, the distance between the two points gets smaller. The closer the points, the closer the line joining them will be to the tangent line.

To differentiate y = x² - x + 3 using first principles, substitute f(x + h) and f(x) into the formula:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[((x+h)^2-(x+h)+3-(x^2-x+3))/((x+h)-x)\right]`

Simplify the numerator:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(x^2+2hx+h^2-x-h+3-x^2+x-3))/(x+h-x)\right]`

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(2hx+h^2-h)/(h)\right]`

Separate into three fractions:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(2hx)/(h)+(h^2)/(h)-(h)/(h)\right]`

Cancel the common factor, h:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[\:2x+h-1\:\right]`

As h → 0, the second term → 0:

`\implies \frac{\text{d}y}{\text{d}x}=2x-1`