Question

Find from the first principle, the derivative with respect to x of the function.y=2x^2-x+3​

Answer 1

`\frac{\text{d}y}{\text{d}x}=4x-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 = 2x² - 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[(2(x+h)^2-(x+h)+3-(2x^2-x+3))/((x+h)-x)\right]`

Simplify the numerator:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(2x^2+4xh+2h^2-x-h+3-2x^2+x-3))/(x+h-x)\right]`

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(2x^2-2x^2+x-x+3-3+4xh+2h^2-h))/(h)\right]`

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(4xh+2h^2-h))/(h)\right]`

Separate into three fractions:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[(4xh)/(h)+(2h^2)/(h)-(h)/(h)\right]`

Cancel the common factor, h:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=\lim_(h \to 0) \left[4x+2h-1\right]`

As h → 0, the second term → 0:

`\implies \displaystyle \frac{\text{d}y}{\text{d}x}=4x-1`