Derivatives 102

Note: $\cfrac{dy}{dx} \, \stackrel{\mathclap{\mbox{def}}}{=} \, \lim\limits_{h\to 0}\cfrac{y(x+h)-y(x)}{h}$

We can work out the derivative of $y = x^2$ by using the definition: $\begin{aligned} \frac{dy}{dx} &= \lim\limits_{h\to 0}\frac{(x+h)^2-x^2}{h} \\ &= \lim\limits_{h\to 0}\frac{x^2 + 2xh + h^2 -x^2}{h} \\ &= \lim\limits_{h\to 0}\frac{2xh + h^2}{h} \\ &= \lim\limits_{h\to 0}\frac{h(2x + h)}{h} \\ &= \lim\limits_{h\to 0}2x + h \\ &= \boxed{2x} \end{aligned}$

Now, we can plug in $-5$ and see that the gradient of the line $y = x^2$ at $x=-5$ is $\huge 2(-5) = \boxed{-10}$