Last week we saw that a derivative is the same thing as the slope of a tangent line. We also saw that the slope to of the tangent line is the same as:
$$\frac{\text{difference in}y}{\text{difference in}x}=\frac{dy}{dx}=\underset{da\to 0}{lim}\frac{f(a+da)-f(a)}{da}$$
But how can this help us find the slope of curves like x² and x³?
x² |
x³ |
Well , let's try looking for the derivative of x² using $\underset{da\to 0}{lim}\frac{f(a+da)-f\left(a\right)}{da}$and see what happens.
Because we are looking for x², the expression should look as follows:
$$\underset{da\to 0}{lim}\frac{(a+da{)}^{2}-{a}^{2}}{da}$$
Now let's start solving!
$$\underset{da\to 0}{lim}\frac{(a+da{)}^{2}-{a}^{2}}{da}=\underset{da\to 0}{lim}\frac{{a}^{2}+2a\mathrm{\u2022\; d}a+d{a}^{2}-a}{da}$$
$$$$$$$$$$=\underset{da\to 0}{lim}\frac{2a\mathrm{\u2022\; d}a+d{a}^{2}}{da}$$
Dividing the top and bottom by "da"should then yield: $\underset{da\to 0}{lim}2a+da$
This then means that the slope of the tangent line is 2a!
Now let's think about what this means. Using the procedure above, we could have found the slope of ANY tangent on the curve. Furthermore, the slope to each and every tangent would always be 2a. In other words, we found more than the slope of some tangent line
We found the slope of the curve!