How do you find the derivative of h(t)=sqrt(t)(1-t^2)

Mathematics

1 answer:

Harrizon [31]3 years ago

5 0

Answer:

Step-by-step explanation:

$h(t)=\sqrt{(t)(1-t^2)}\\h^{2}(t)=t(1-t^2)=t-t^3\\ diff. w.r.t.,\\2 h(t)h'(t)=1-3t^2\\h'(t)=\frac{1-3t^2}{2 h(t)} \\h'(t)=\frac{1-3t^2}{2\sqrt{(t)(1-t^2)} }$

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1. Complete the inequality below to describe the

dezoksy [38]

Domain is the entire span left to right (on the x-axis) that the graph is on. Since the graph goes from x=-4 and ends at x=4, the domain would be from -4 to 4. The circle at -4 is open, so it does not include the point at -4, just everything leading up to it. So, the domain would be

$- 4 < x \leqslant 4$

The range is similar, it is the entire span that the graph goes up and down (on the y-axis). The graph starts at the bottom at y=-2, and ends at y=5. The bottom point (4,-2) is closed, so the graph includes that point, and the top point (-4,5) is open and doesn't include the point. Therefore, the range would be

$- 2 \leqslant x < 5$