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# Tutor profile: Theo B.

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Theo B.
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## Questions

### Subject:Calculus

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Question:

What is the derivative of $$\tan x$$?

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Theo B.

Since $$\tan{x} = \frac{\sin x}{\cos x}$$ we will use the quotient rule. The quotient rule states that $$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{gf' - fg'}{g^2}$$. So $$\frac{d}{dx}\tan{x} = \frac{(\cos{x} )(\cos{x}) - (\sin{x})(-\sin{x}) }{\cos{^2 x}} = \frac{\cos{^2x} + \sin{^2x} }{\cos{^2 x}}$$. So using the trigonomic identity $$\cos{^2x} + \sin{^2x} = 1$$, it must be that $$\frac{d}{dx}\tan{x} =\frac{1}{\cos{^2 x}} = \sec{^2x}$$.

### Subject:Physics

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Question:

If a projectile is dropped from rest at a height $$h$$, what is its velocity right before it contacts the ground?

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Theo B.

The vertical displacement of the projectile is $$y = h - \frac{gt^2}{2}$$, and its velocity is $$v = -gt$$ (from equations of motion for a projectile). Where $$t$$ is the time elapsed and $$g = 9.8 \frac{m}{s^2}$$ is the acceleration due to gravity. We need to know when the projectile hits the ground to figure out its velocity at impact. At impact, the projectile's vertical displacement will be zero. We then must have $$y = 0$$ at impact, so $$h = \frac{gt^2}{2}$$ and thus $$t = \sqrt{\frac{2h}{g}}$$. Plugging this result into the velocity gives an impact velocity of $$v = -g\sqrt{\frac{2h}{g}} = -\sqrt{2hg}$$.

### Subject:Algebra

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Question:

What values of $$x$$ make $$x^2 - 4x + 3 > 0$$?

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Theo B.

First we factor the expression like so $$x^2 - 4x + 3 = (x - 3)(x - 1)$$. Then we note that the expression's sign can only change at its zeros. Thus we must consider the three intervals $$x < 1$$, $$1 < x < 3$$, and $$x > 3$$. If $$x < 1$$, then $$x - 3 < - 2$$ and $$x - 1 < 0$$ so $$(x - 3)(x - 1) > 0$$. If $$1 < x < 3$$, then $$x - 1 > 0$$ and $$x - 3 < 0$$ so $$(x - 3)(x - 1) < 0$$. Finally if $$x > 3$$, then $$x - 1 > 0$$ and $$x - 3 > 0$$ so $$(x - 3)(x - 1) > 0$$. Therefore the answer is that $$x^2 - 4x + 3 > 0$$ when $$x < 1$$ or $$x > 3$$.

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