Differentiation from first principles

Question

How would I prove that

$$\frac d{dt}(1+t-2t^2) = 1-4t$$

Using differentiation from first principles.

Then I tried to use the equation:

$$\frac{f(t+h)-f(t)}h$$

Is this correct and what do I do after this.

Answer 1

Yes, that’s the First Principle, or the definition of a derivative:

$$f’(t) = \lim_{h \to 0}\frac{f(t+h)-f(t)}{h}$$

You have $f(t) = 1+t-2t^2$, so $f(t+h) = 1+(t+h)-2(t+h)^2$. From here, you need to evaluate

$$\lim_{h \to 0}\frac{\left[1+(t+h)-2(t+h)^2\right]-\left[1+t-2t^2\right]}{h}$$

Simply note that $(a+b)^2 = a^2+2ab+b^2$ which you most probably know. You can solve this simply as the limit simplifies very quickly.

Answer 2

How would I prove that: d/dr (1+t-2t^2) = 1-4t

I assume you want to find the derivative with respect to $t$, not $r$.

Using differentiation from first principles. I tried to integrate the equation and got the following: f(t) =(1t+.5t^2-2/3t^3)

Why would you integrate if you want to differentiate (from first principles or otherwise)...?

Then I tried to uses the equation: f(t+h)-f(t) / h

That's better, use the definition and find the following limit: $$\lim_{h \to 0}\frac{f(t+h)-f(t)}{h}$$ for $f(t) = 1+t-2t^2$.

Is this correct and what do I do after this.

Use $f$ to evaluate $f(t+h)$ and $f(t)$ in the limit above: substitute and simplify first.