Answer
As $h\to0$, the curve $y=\frac{\cos((x+h)^2)-\cos(x^2)}{h}$ gets closer and closer to the curve $y=-2x\sin(x^2)$, apparently approaching to become one with $y=-2x\sin(x^2)$.
Work Step by Step
The graphs of the function are enclosed below. The graph of $y=-2x\sin(x^2)$ is portrayed by the black curve, while the other curves are variations of $y=\frac{\cos((x+h)^2)-\cos(x^2)}{h}$.
From the graph, we see that as $h\to0$, the curve $y=\frac{\cos((x+h)^2)-\cos(x^2)}{h}$ gets closer and closer to the curve $y=-2x\sin(x^2)$, apparently approaching to become one with $y=-2x\sin(x^2)$.
In other words, we can say that $$\lim_{h\to0}\frac{\cos((x+h)^2)-\cos(x^2)}{h}=-2x\sin(x^2)$$
From the definition of derivative, we know that $$\lim_{h\to0}\frac{\cos((x+h)^2)-\cos(x^2)}{h}=(\cos (x^2))'$$
Therefore, $$(\cos(x^2))'=-2x\sin(x^2)$$
So the behavior of the graphs can be summarized to show that the derivative of $\cos(x^2)$ is $-2x\sin(x^2)$.
