3.4-3.5 Review WS featuring multiple-choice questions on calculus concepts including derivatives, tangent lines, inverse functions, and related rates.

Let's solve each problem step-by-step from the worksheet.

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1. Find $ f'\left(\frac{\pi}{8}\right) $ if $ f(x) = \sec^2(2x) $.

We are given:
$$
f(x) = \sec^2(2x)
$$

This is a composite function, so we use the chain rule.

Let’s differentiate:
$$
f'(x) = 2\sec(2x) \cdot \frac{d}{dx}[\sec(2x)] = 2\sec(2x) \cdot \sec(2x)\tan(2x) \cdot 2
$$
$$
= 4\sec^2(2x)\tan(2x)
$$

Now evaluate at $ x = \frac{\pi}{8} $:

- $ 2x = \frac{\pi}{4} $
- $ \sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos(\pi/4)} = \frac{1}{\sqrt{2}/2} = \sqrt{2} $
- So $ \sec^2(\pi/4) = (\sqrt{2})^2 = 2 $
- $ \tan(\pi/4) = 1 $

So:
$$
f'\left(\frac{\pi}{8}\right) = 4 \cdot 2 \cdot 1 = 8
$$

✔ Answer: e. 8

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2. Find the tangent line to the ellipse $ \frac{x^2}{16} + \frac{y^2}{4} = 1 $ at the point $ (2, -\sqrt{3}) $.

Use implicit differentiation:

Differentiate both sides:
$$
\frac{2x}{16} + \frac{2y y'}{4} = 0 \Rightarrow \frac{x}{8} + \frac{y y'}{2} = 0
$$

Solve for $ y' $:
$$
\frac{y y'}{2} = -\frac{x}{8} \Rightarrow y' = -\frac{x}{8} \cdot \frac{2}{y} = -\frac{x}{4y}
$$

At $ (2, -\sqrt{3}) $:
$$
y' = -\frac{2}{4(-\sqrt{3})} = -\frac{2}{-4\sqrt{3}} = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6}
$$

Now use point-slope form:
$$
y + \sqrt{3} = \frac{\sqrt{3}}{6}(x - 2)
$$

Simplify:
$$
y = \frac{\sqrt{3}}{6}x - \frac{2\sqrt{3}}{6} - \sqrt{3} = \frac{\sqrt{3}}{6}x - \frac{\sqrt{3}}{3} - \sqrt{3}
= \frac{\sqrt{3}}{6}x - \frac{4\sqrt{3}}{3}
$$

But this doesn't match any option exactly. Let's check options:

a. $ y = -\sqrt{6}x - 3 $
b. none of these
c. $ y = \sqrt{3}x - 4 $
d. $ y = \frac{\sqrt{3}}{6}x - \frac{4\sqrt{3}}{3} $ ← matches our result!
e. $ y = \frac{\sqrt{3}}{3}x - 4 $

Wait — option d is exactly what we got.

✔ Answer: d. $ y = \frac{\sqrt{3}}{6}x - \frac{4\sqrt{3}}{3} $

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3. Find the derivative of $ y = 2\cos^{-1}(\sin^{-1}(t)) $

Let:
$$
y = 2\cos^{-1}(u), \quad u = \sin^{-1}(t)
$$

Then:
$$
\frac{dy}{dt} = 2 \cdot \left( -\frac{1}{\sqrt{1 - u^2}} \right) \cdot \frac{du}{dt}
= -\frac{2}{\sqrt{1 - (\sin^{-1}(t))^2}} \cdot \frac{1}{\sqrt{1 - t^2}}
$$

So:
$$
y' = -\frac{2}{\sqrt{1 - (\sin^{-1}(t))^2} \cdot \sqrt{1 - t^2}}
$$

Compare with options:

a. $ y' = -\frac{2}{\sqrt{(1+t^2)\left[1 - (\sin^{-1}(t))^2\right]}} $ → wrong
b. $ y' = -\frac{2}{\sqrt{1 - t^2}} $ → missing denominator
c. $ y' = -\frac{2}{\sqrt{1 - (\sin^{-1}(t))^2}} $ → missing $ \sqrt{1 - t^2} $
d. $ y' = -\frac{2}{\sqrt{(1 - t^2)\left[1 - \sin^{-1}(t)\right]}} $ → wrong
e. $ y' = -\frac{2}{\sqrt{(1 - t^2)\left[1 - (\sin^{-1}(t))^2\right]}} $ ✔

Yes, this matches our expression.

✔ Answer: e

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4. If $ h $ is the inverse of $ f $ and $ f(x) = \frac{1}{x} $, find $ h'(3) $.

Given $ f(x) = \frac{1}{x} $, then its inverse is $ h(x) = \frac{1}{x} $ as well, since $ f(f(x)) = x $.

But more generally:
$$
h'(x) = \frac{1}{f'(h(x))}
$$

We want $ h'(3) $. First, find $ h(3) $: since $ f(x) = 1/x $, then $ h(x) = 1/x $, so $ h(3) = 1/3 $

Now $ f'(x) = -1/x^2 $, so $ f'(1/3) = -1/(1/9) = -9 $

Thus:
$$
h'(3) = \frac{1}{f'(h(3))} = \frac{1}{f'(1/3)} = \frac{1}{-9} = -\frac{1}{9}
$$

Wait — but that's not among the options? Wait — let's recheck.

Actually, options are not listed here in the image. But assuming they are standard multiple choice, perhaps I missed something.

Wait — the question says "find $ h'(3) $" and no options are shown in your image. But likely it's a numerical value.

But let's double-check.

We know:
- $ f(x) = 1/x $
- So $ f^{-1}(x) = 1/x $ → $ h(x) = 1/x $
- Then $ h'(x) = -1/x^2 $
- So $ h'(3) = -1/9 $

But again, if the choices aren’t visible, maybe it’s implied. But based on logic, answer is $ -\frac{1}{9} $

But wait — perhaps the question is asking for $ h'(3) $, and we need to use formula:
$$
h'(x) = \frac{1}{f'(h(x))}
$$

We already did that.

Alternatively, since $ f(x) = 1/x $, then $ f(a) = b $ means $ a = 1/b $

So $ f(1/3) = 3 $, so $ h(3) = 1/3 $

$ f'(x) = -1/x^2 $, so $ f'(1/3) = -9 $

Then $ h'(3) = 1 / (-9) = -1/9 $

✔ So $ h'(3) = -\frac{1}{9} $

But since options aren't shown, we’ll move on.

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5. Find $ y' $ if $ y = \tan^{-1}\left(\frac{x}{2}\right) $

Use chain rule:
$$
\frac{dy}{dx} = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2}
= \frac{1}{1 + \frac{x^2}{4}} \cdot \frac{1}{2}
= \frac{1}{\frac{4 + x^2}{4}} \cdot \frac{1}{2}
= \frac{4}{4 + x^2} \cdot \frac{1}{2} = \frac{2}{4 + x^2}
$$

So $ y' = \frac{2}{x^2 + 4} $

No options shown, but that’s the answer.

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6. Find $ \frac{dy}{dx} $ if $ y = \sin^{-1}x - \sqrt{1 - x^2} $

Differentiate term by term:

- $ \frac{d}{dx}[\sin^{-1}x] = \frac{1}{\sqrt{1 - x^2}} $
- $ \frac{d}{dx}[\sqrt{1 - x^2}] = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} $

So:
$$
\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} - \left( \frac{-x}{\sqrt{1 - x^2}} \right) = \frac{1 + x}{\sqrt{1 - x^2}}
$$

Wait — no:

Wait, $ y = \sin^{-1}x - \sqrt{1 - x^2} $

So derivative:
$$
\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} - \left( \frac{-x}{\sqrt{1 - x^2}} \right) = \frac{1}{\sqrt{1 - x^2}} + \frac{x}{\sqrt{1 - x^2}} = \frac{1 + x}{\sqrt{1 - x^2}}
$$

✔ Answer: $ \frac{1 + x}{\sqrt{1 - x^2}} $

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7. Which equation is perpendicular to the tangent line to $ f(x) = \cos^{-1}x $ at $ x = \frac{\sqrt{3}}{2} $?

First, find slope of tangent line at $ x = \frac{\sqrt{3}}{2} $

We know:
$$
f'(x) = \frac{d}{dx}[\cos^{-1}x] = -\frac{1}{\sqrt{1 - x^2}}
$$

At $ x = \frac{\sqrt{3}}{2} $:
$$
f'\left(\frac{\sqrt{3}}{2}\right) = -\frac{1}{\sqrt{1 - \left(\frac{3}{4}\right)}} = -\frac{1}{\sqrt{1/4}} = -\frac{1}{1/2} = -2
$$

So slope of tangent is $ -2 $

Then slope of perpendicular line is $ \frac{1}{2} $ (negative reciprocal)

Now find $ y $-value at $ x = \frac{\sqrt{3}}{2} $:
$$
f\left(\frac{\sqrt{3}}{2}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}
$$

So point is $ \left( \frac{\sqrt{3}}{2}, \frac{\pi}{6} \right) $

Now find line through this point with slope $ \frac{1}{2} $:

$$
y - \frac{\pi}{6} = \frac{1}{2}\left(x - \frac{\sqrt{3}}{2}\right)
$$

Check options:

a. $ y - \frac{\pi}{6} = -2\left(x - \frac{\sqrt{3}}{2}\right) $ → slope -2 → tangent, not perpendicular
b. $ y - \frac{\pi}{3} = -2(...) $ → wrong y-value
c. $ y - \frac{\pi}{6} = -\frac{1}{2}(x - \frac{\sqrt{3}}{2}) $ → slope -1/2 → not correct
d. $ y - \frac{\pi}{3} = \frac{1}{2}(x - \frac{\sqrt{3}}{2}) $ → wrong y-value
e. $ y - \frac{\pi}{6} = \frac{1}{2}(x - \frac{\sqrt{3}}{2}) $ ✔

Yes! This matches.

✔ Answer: e

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8. Given lens equation: $ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $, $ f $ constant. Find $ \frac{dv}{du} $

Differentiate both sides w.r.t. $ u $, treating $ v $ as function of $ u $:

$$
0 = -\frac{1}{v^2} \frac{dv}{du} - \frac{1}{u^2}
$$

So:
$$
\frac{1}{v^2} \frac{dv}{du} = -\frac{1}{u^2}
\Rightarrow \frac{dv}{du} = -\frac{v^2}{u^2}
$$

But we want in terms of $ u $ and $ f $. Recall:
$$
\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{uf}
\Rightarrow v = \frac{uf}{u - f}
$$

So $ v = \frac{uf}{u - f} $

Then:
$$
v^2 = \left( \frac{uf}{u - f} \right)^2
\Rightarrow \frac{dv}{du} = -\frac{v^2}{u^2} = -\frac{1}{u^2} \cdot \left( \frac{u^2 f^2}{(u - f)^2} \right) = -\frac{f^2}{(u - f)^2}
$$

✔ Answer: d. $ \frac{dv}{du} = -\frac{f^2}{(u - f)^2} $

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9. Find tangent to $ y = \sqrt{1 + 4\sin x} $ at $ (0, 1) $

First, verify point: $ x = 0 $, $ y = \sqrt{1 + 0} = 1 $ → yes.

Now derivative:
$$
\frac{dy}{dx} = \frac{1}{2\sqrt{1 + 4\sin x}} \cdot (4\cos x) = \frac{2\cos x}{\sqrt{1 + 4\sin x}}
$$

At $ x = 0 $:
$$
\frac{dy}{dx} = \frac{2 \cdot 1}{\sqrt{1 + 0}} = 2
$$

So slope = 2, point $ (0,1) $

Equation: $ y - 1 = 2(x - 0) \Rightarrow y = 2x + 1 $

✔ Answer: e. $ y = 2x + 1 $

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10. Suppose $ F(x) = f(g(x)) $, $ g(10) = 3 $, $ g'(10) = 6 $, $ f'(10) = 7 $, $ f'(3) = 9 $. Find $ F'(10) $

Chain rule:
$$
F'(x) = f'(g(x)) \cdot g'(x)
\Rightarrow F'(10) = f'(g(10)) \cdot g'(10) = f'(3) \cdot 6 = 9 \cdot 6 = 54
$$

✔ Answer: b. 54

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11. If $ y = \frac{2}{(3x - 4)^4} $, find $ \frac{dy}{dx} $

Rewrite: $ y = 2(3x - 4)^{-4} $

Differentiate:
$$
\frac{dy}{dx} = 2 \cdot (-4)(3x - 4)^{-5} \cdot 3 = -24(3x - 4)^{-5}
= \frac{-24}{(3x - 4)^5}
$$

✔ Answer: c. $ \frac{-24}{(3x - 4)^5} $

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✔ Final Answers Summary:

1. e. 8
2. d. $ y = \frac{\sqrt{3}}{6}x - \frac{4\sqrt{3}}{3} $
3. e. $ y' = -\frac{2}{\sqrt{(1 - t^2)\left[1 - (\sin^{-1}(t))^2\right]}} $
4. $ h'(3) = -\frac{1}{9} $ (not in options?)
5. $ y' = \frac{2}{x^2 + 4} $
6. $ \frac{dy}{dx} = \frac{1 + x}{\sqrt{1 - x^2}} $
7. e. $ y - \frac{\pi}{6} = \frac{1}{2}\left(x - \frac{\sqrt{3}}{2}\right) $
8. d. $ \frac{dv}{du} = -\frac{f^2}{(u - f)^2} $
9. e. $ y = 2x + 1 $
10. b. 54
11. c. $ \frac{-24}{(3x - 4)^5} $

Let me know if you'd like explanations in a different format or need help with anything else!