Inverse Functions Worksheet for 12th - Higher Ed | Lesson Planet - Free Printable

Since I can't view or access images directly, I can't see the specific content of your uploaded image. However, based on your description, it appears to be a Calculus Review Worksheet with several problems involving derivatives, implicit differentiation, and tangent lines.

Let me help you solve each problem step-by-step, assuming the worksheet contains the following typical calculus problems (as inferred from your text):

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(1) Find $\frac{dy}{dx}$ implicitly:

We are given equations involving $x$ and $y$, and we need to find $\frac{dy}{dx}$ using implicit differentiation.

#### (a) $ x = \sin^{-1}(\sqrt{x^2 + y^2}) $

This is tricky because $x$ is expressed in terms of an inverse trig function involving both $x$ and $y$. Let's differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

$$
\frac{d}{dx}[x] = \frac{d}{dx}\left[\sin^{-1}(\sqrt{x^2 + y^2})\right]
$$

Left side: $1$

Right side: Use chain rule.

Let $u = \sqrt{x^2 + y^2} = (x^2 + y^2)^{1/2}$

Then:
$$
\frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}
$$

Now compute $\frac{du}{dx}$:

$$
\frac{du}{dx} = \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot (2x + 2y \frac{dy}{dx}) = \frac{x + y \frac{dy}{dx}}{\sqrt{x^2 + y^2}}
$$

So,
$$
1 = \frac{1}{\sqrt{1 - (\sqrt{x^2 + y^2})^2}} \cdot \frac{x + y \frac{dy}{dx}}{\sqrt{x^2 + y^2}}
$$

Note: $(\sqrt{x^2 + y^2})^2 = x^2 + y^2$, so:

$$
1 = \frac{1}{\sqrt{1 - (x^2 + y^2)}} \cdot \frac{x + y \frac{dy}{dx}}{\sqrt{x^2 + y^2}}
$$

Multiply both sides by $\sqrt{1 - (x^2 + y^2)} \cdot \sqrt{x^2 + y^2}$:

$$
\sqrt{1 - (x^2 + y^2)} \cdot \sqrt{x^2 + y^2} = x + y \frac{dy}{dx}
$$

Solve for $\frac{dy}{dx}$:

$$
y \frac{dy}{dx} = \sqrt{1 - (x^2 + y^2)} \cdot \sqrt{x^2 + y^2} - x
$$

$$
\frac{dy}{dx} = \frac{ \sqrt{(1 - x^2 - y^2)(x^2 + y^2)} - x }{y}
$$

✔ This is the derivative.

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#### (b) $ y = x^{e^x} $

This is exponential function with variable base and exponent. Use logarithmic differentiation.

Take natural log of both sides:

$$
\ln y = \ln(x^{e^x}) = e^x \ln x
$$

Differentiate both sides:

$$
\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(e^x \ln x) = e^x \ln x + e^x \cdot \frac{1}{x} = e^x \left( \ln x + \frac{1}{x} \right)
$$

So:

$$
\frac{dy}{dx} = y \cdot e^x \left( \ln x + \frac{1}{x} \right) = x^{e^x} \cdot e^x \left( \ln x + \frac{1}{x} \right)
$$

✔ Final answer:
$$
\boxed{\frac{dy}{dx} = x^{e^x} e^x \left( \ln x + \frac{1}{x} \right)}
$$

---

#### (c) $ x = \cos^{-1}(y^3) $

Differentiate both sides w.r.t. $x$:

$$
\frac{d}{dx}[x] = \frac{d}{dx}[\cos^{-1}(y^3)]
$$

Left: $1$

Right: Use chain rule:

$$
\frac{d}{dx}[\cos^{-1}(y^3)] = -\frac{1}{\sqrt{1 - (y^3)^2}} \cdot \frac{d}{dx}(y^3) = -\frac{1}{\sqrt{1 - y^6}} \cdot 3y^2 \frac{dy}{dx}
$$

So:

$$
1 = -\frac{3y^2}{\sqrt{1 - y^6}} \frac{dy}{dx}
$$

Solve for $\frac{dy}{dx}$:

$$
\frac{dy}{dx} = -\frac{\sqrt{1 - y^6}}{3y^2}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = -\frac{\sqrt{1 - y^6}}{3y^2}}
$$

---

#### (d) $ y = \tan^{-1}(\sqrt{x^2 + 1}) $

Use chain rule:

Let $u = \sqrt{x^2 + 1} = (x^2 + 1)^{1/2}$

Then:
$$
\frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx}
$$

Compute $\frac{du}{dx}$:

$$
\frac{du}{dx} = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}
$$

Now:
$$
\frac{dy}{dx} = \frac{1}{1 + (x^2 + 1)} \cdot \frac{x}{\sqrt{x^2 + 1}} = \frac{1}{x^2 + 2} \cdot \frac{x}{\sqrt{x^2 + 1}}
$$

✔ Final answer:
$$
\boxed{\frac{dy}{dx} = \frac{x}{(x^2 + 2)\sqrt{x^2 + 1}}}
$$

---

#### (e) $ y = \ln(\ln x) $

Chain rule:

$$
\frac{dy}{dx} = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = \frac{1}{x \ln x}}
$$

---

#### (f) $ y = \sin^{-1}(\sqrt{x^2 + y^2}) $

Wait — this looks like part (a), but now $y$ is on the left. Let's differentiate:

$$
\frac{dy}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x^2 + y^2})^2}} \cdot \frac{d}{dx}(\sqrt{x^2 + y^2})
$$

As before, $\sqrt{x^2 + y^2}$ has derivative:
$$
\frac{x + y \frac{dy}{dx}}{\sqrt{x^2 + y^2}}
$$

So:

$$
\frac{dy}{dx} = \frac{1}{\sqrt{1 - (x^2 + y^2)}} \cdot \frac{x + y \frac{dy}{dx}}{\sqrt{x^2 + y^2}}
$$

Multiply both sides by $\sqrt{1 - (x^2 + y^2)} \cdot \sqrt{x^2 + y^2}$:

$$
\frac{dy}{dx} \cdot \sqrt{1 - (x^2 + y^2)} \cdot \sqrt{x^2 + y^2} = x + y \frac{dy}{dx}
$$

Bring all terms to one side:

$$
\frac{dy}{dx} \left[ \sqrt{(1 - x^2 - y^2)(x^2 + y^2)} - y \right] = x
$$

So:

$$
\frac{dy}{dx} = \frac{x}{\sqrt{(1 - x^2 - y^2)(x^2 + y^2)} - y}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = \frac{x}{\sqrt{(1 - x^2 - y^2)(x^2 + y^2)} - y}}
$$

---

#### (g) $ y = \ln^2(\sqrt{x}) $

Note: $\ln^2(\sqrt{x}) = [\ln(\sqrt{x})]^2 = \left( \frac{1}{2} \ln x \right)^2 = \frac{1}{4} (\ln x)^2$

So:

$$
\frac{dy}{dx} = \frac{1}{4} \cdot 2 \ln x \cdot \frac{1}{x} = \frac{\ln x}{2x}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = \frac{\ln x}{2x}}
$$

---

#### (h) $ y = \frac{1}{\sqrt{x^2 + 1}} $

Rewrite: $ y = (x^2 + 1)^{-1/2} $

Differentiate:

$$
\frac{dy}{dx} = -\frac{1}{2}(x^2 + 1)^{-3/2} \cdot 2x = -\frac{x}{(x^2 + 1)^{3/2}}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = -\frac{x}{(x^2 + 1)^{3/2}}}
$$

---

#### (i) $ y = \sin(2x) \cdot \cos(3x) $

Product rule:

$$
\frac{dy}{dx} = \cos(2x) \cdot 2 \cdot \cos(3x) + \sin(2x) \cdot (-\sin(3x)) \cdot 3
$$

$$
= 2 \cos(2x)\cos(3x) - 3 \sin(2x)\sin(3x)
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = 2 \cos(2x)\cos(3x) - 3 \sin(2x)\sin(3x)}
$$

---

#### (j) $ y = \tan(x) \cdot \sec(x) $

Product rule:

$$
\frac{dy}{dx} = \sec^2(x) \cdot \sec(x) + \tan(x) \cdot \sec(x)\tan(x)
= \sec^3(x) + \tan^2(x)\sec(x)
$$

Factor $\sec(x)$:

$$
= \sec(x)(\sec^2(x) + \tan^2(x))
$$

But $\sec^2(x) = 1 + \tan^2(x)$, so:

$$
\sec^2(x) + \tan^2(x) = 1 + 2\tan^2(x)
$$

Alternatively, leave as:

$$
\boxed{\frac{dy}{dx} = \sec^3(x) + \sec(x)\tan^2(x)}
$$

Or factor:
$$
\boxed{\frac{dy}{dx} = \sec(x)(\sec^2(x) + \tan^2(x))}
$$

---

#### (k) $ y = \sin(e^x) $

Chain rule:

$$
\frac{dy}{dx} = \cos(e^x) \cdot e^x = e^x \cos(e^x)
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = e^x \cos(e^x)}
$$

---

#### (l) $ y = \ln(\sin(x)) $

$$
\frac{dy}{dx} = \frac{1}{\sin(x)} \cdot \cos(x) = \cot(x)
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = \cot(x)}
$$

---

#### (m) $ y = \cos^{-1}(x) $

Standard derivative:

$$
\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}}
$$

---

#### (n) $ y = \frac{\sin(x)}{x^2 + 1} $

Quotient rule:

Numerator: $u = \sin(x)$, $u' = \cos(x)$

Denominator: $v = x^2 + 1$, $v' = 2x$

$$
\frac{dy}{dx} = \frac{\cos(x)(x^2 + 1) - \sin(x)(2x)}{(x^2 + 1)^2}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = \frac{(x^2 + 1)\cos(x) - 2x \sin(x)}{(x^2 + 1)^2}}
$$

---

#### (o) $ y = \arctan(x^2) $

$$
\frac{dy}{dx} = \frac{1}{1 + (x^2)^2} \cdot 2x = \frac{2x}{1 + x^4}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = \frac{2x}{1 + x^4}}
$$

---

#### (p) $ y = \ln(x^2 + 1) $

$$
\frac{dy}{dx} = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}
$$

✔ Answer:
$$
\boxed{\frac{dy}{dx} = \frac{2x}{x^2 + 1}}
$$

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(2) Let $ f(x) = x^2 + 4x $, $ x > -2 $, and let $ g $ be the inverse of $ f $. Find $ g'(x) $ when $ x = 4 $.

We use the formula:
$$
g'(x) = \frac{1}{f'(g(x))}
$$

First, find $f'(x)$:
$$
f'(x) = 2x + 4
$$

We want $g'(4)$, so we need $g(4)$, i.e., the value of $x$ such that $f(x) = 4$.

Solve:
$$
x^2 + 4x = 4 \Rightarrow x^2 + 4x - 4 = 0
$$

Use quadratic formula:
$$
x = \frac{-4 \pm \sqrt{16 + 16}}{2} = \frac{-4 \pm \sqrt{32}}{2} = \frac{-4 \pm 4\sqrt{2}}{2} = -2 \pm 2\sqrt{2}
$$

Since $x > -2$, we take $x = -2 + 2\sqrt{2}$

So $g(4) = -2 + 2\sqrt{2}$

Now compute $f'(g(4)) = f'(-2 + 2\sqrt{2}) = 2(-2 + 2\sqrt{2}) + 4 = -4 + 4\sqrt{2} + 4 = 4\sqrt{2}$

Thus:
$$
g'(4) = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}
$$

✔ Final answer:
$$
\boxed{g'(4) = \frac{\sqrt{2}}{8}}
$$

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(3) Write the equation of the line which is tangent to each of the following:

#### (a) $ y = x^2 $ at $ x = 3 $

Find point: $y = 9$, so point is $(3, 9)$

Derivative: $y' = 2x$, so slope at $x = 3$: $6$

Equation: $y - 9 = 6(x - 3)$ → $y = 6x - 9$

✔ Answer:
$$
\boxed{y = 6x - 9}
$$

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#### (b) $ y = \ln(x) $ at $ x = 1 $

Point: $y = \ln(1) = 0$, so $(1, 0)$

Derivative: $y' = \frac{1}{x}$, so slope at $x = 1$: $1$

Equation: $y - 0 = 1(x - 1)$ → $y = x - 1$

✔ Answer:
$$
\boxed{y = x - 1}
$$

---

#### (c) $ y = \tan^{-1}(x) $ at $ x = 0 $

Point: $y = \tan^{-1}(0) = 0$, so $(0, 0)$

Derivative: $y' = \frac{1}{1 + x^2}$, so at $x = 0$: $1$

Equation: $y = x$

✔ Answer:
$$
\boxed{y = x}
$$

---

#### (d) $ \tan^{-1}(y) = \cos^{-1}(2x) $ at $ (1, \pi/2) $

Check if point lies on curve:

Left: $\tan^{-1}(\pi/2) \approx \tan^{-1}(1.57) \approx 1.0$

Right: $\cos^{-1}(2 \cdot 1) = \cos^{-1}(2)$ → undefined! Wait.

$\cos^{-1}(2)$ is not defined since domain of $\cos^{-1}$ is $[-1, 1]$.

So $x = 1$ → $2x = 2$, not valid.

Wait — maybe typo? Or perhaps $x = 0.5$?

Try $x = 0.5$: $2x = 1$, $\cos^{-1}(1) = 0$

Then $\tan^{-1}(y) = 0$ → $y = 0$

So point $(0.5, 0)$

But given point is $(1, \pi/2)$ — not valid.

Wait: could it be $ \tan^{-1}(y) = \cos^{-1}(2x) $, and we're told the point is $(1, \pi/2)$?

At $x = 1$: $2x = 2$, $\cos^{-1}(2)$ is undefined.

So likely a typo.

Maybe it’s $ \tan^{-1}(y) = \cos^{-1}(x) $, or $2x$ should be $x/2$?

Alternatively, suppose the point is $(0, \pi/2)$?

At $x = 0$: $\cos^{-1}(0) = \pi/2$, so $\tan^{-1}(y) = \pi/2$ → $y \to \infty$, not possible.

Wait: $\tan^{-1}(y) = \pi/2$ only as $y \to \infty$, never reached.

So no point satisfies $\tan^{-1}(y) = \cos^{-1}(2x)$ with $y = \pi/2$

Because $\tan^{-1}(y) < \pi/2$ always.

So $\tan^{-1}(y) = \pi/2$ is impossible.

Therefore, the point $(1, \pi/2)$ does not lie on the curve unless there's a typo.

Possibility: Maybe it's $ \tan^{-1}(y) = \cos^{-1}(x) $, and point is $(0, \pi/2)$?

At $x = 0$: $\cos^{-1}(0) = \pi/2$, so $\tan^{-1}(y) = \pi/2$ → again, $y \to \infty$, not possible.

Wait: maybe the equation is $ \tan^{-1}(y) = \cos^{-1}(2x) $, and the point is $(0.5, 0)$?

At $x = 0.5$: $2x = 1$, $\cos^{-1}(1) = 0$, so $\tan^{-1}(y) = 0$ → $y = 0$

So point $(0.5, 0)$

But the problem says $(1, \pi/2)$ — likely a typo.

Alternatively, perhaps the equation is $ \tan^{-1}(y) = \cos^{-1}(x) $, and point is $(0, \pi/2)$?

Still invalid.

Wait — maybe it's $ \tan^{-1}(y) = \cos^{-1}(x) $, and we're to find tangent at $x = 0$, $y = \pi/2$? But $y = \pi/2$ gives $\tan^{-1}(y) \approx 1.57$, $\cos^{-1}(0) = \pi/2 \approx 1.57$, so yes!

So if $ \tan^{-1}(y) = \cos^{-1}(x) $, then at $x = 0$, $y = \pi/2$ is a solution.

So assume the equation is:
$$
\tan^{-1}(y) = \cos^{-1}(x)
$$
and point $(0, \pi/2)$

Differentiate implicitly:

Left: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx}$

Right: $-\frac{1}{\sqrt{1 - x^2}}$

So:
$$
\frac{1}{1 + y^2} \cdot \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}
$$

At $(0, \pi/2)$:

Left: $\frac{1}{1 + (\pi/2)^2} \cdot \frac{dy}{dx}$

Right: $-\frac{1}{\sqrt{1 - 0}} = -1$

So:
$$
\frac{dy}{dx} = -1 \cdot (1 + (\pi/2)^2) = -(1 + \pi^2/4)
$$

Then tangent line:

$$
y - \frac{\pi}{2} = m(x - 0), \quad m = -\left(1 + \frac{\pi^2}{4}\right)
$$

So:
$$
y = -\left(1 + \frac{\pi^2}{4}\right)x + \frac{\pi}{2}
$$

But this is speculative due to possible typo.

Alternatively, if original was correct, perhaps $x = 0.5$, $y = 0$?

Let’s assume the intended point is $(0.5, 0)$ for $ \tan^{-1}(y) = \cos^{-1}(2x) $

At $x = 0.5$: $2x = 1$, $\cos^{-1}(1) = 0$, so $\tan^{-1}(y) = 0$ → $y = 0$

So point $(0.5, 0)$

Differentiate:

$$
\frac{1}{1 + y^2} \cdot \frac{dy}{dx} = -\frac{2}{\sqrt{1 - (2x)^2}} = -\frac{2}{\sqrt{1 - 4x^2}}
$$

At $x = 0.5$, $y = 0$:

Left: $\frac{1}{1 + 0} \cdot \frac{dy}{dx} = \frac{dy}{dx}$

Right: $-\frac{2}{\sqrt{1 - 4(0.25)}} = -\frac{2}{\sqrt{1 - 1}} = \text{undefined!}$

So vertical tangent? Not possible.

So perhaps the original equation is different.

Given the ambiguity, and since the point $(1, \pi/2)$ cannot satisfy the equation, I suspect a typo.

Perhaps it's $ \tan^{-1}(y) = \cos^{-1}(x) $, and point $(0, \pi/2)$ — but still invalid.

Another idea: maybe $ y = \tan^{-1}(x) $, and point $(1, \pi/4)$? But that's not what's written.

Given time, and since the problem is ambiguous, I’ll stop here.

But if you can confirm the exact equation and point, I can fix it.

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✔ Summary of Answers:

| Problem | Answer |
|--------|--------|
| (1a) | $\frac{dy}{dx} = \frac{ \sqrt{(1 - x^2 - y^2)(x^2 + y^2)} - x }{y}$ |
| (1b) | $\frac{dy}{dx} = x^{e^x} e^x \left( \ln x + \frac{1}{x} \right)$ |
| (1c) | $\frac{dy}{dx} = -\frac{\sqrt{1 - y^6}}{3y^2}$ |
| (1d) | $\frac{dy}{dx} = \frac{x}{(x^2 + 2)\sqrt{x^2 + 1}}$ |
| (1e) | $\frac{dy}{dx} = \frac{1}{x \ln x}$ |
| (1f) | $\frac{dy}{dx} = \frac{x}{\sqrt{(1 - x^2 - y^2)(x^2 + y^2)} - y}$ |
| (1g) | $\frac{dy}{dx} = \frac{\ln x}{2x}$ |
| (1h) | $\frac{dy}{dx} = -\frac{x}{(x^2 + 1)^{3/2}}$ |
| (1i) | $\frac{dy}{dx} = 2 \cos(2x)\cos(3x) - 3 \sin(2x)\sin(3x)$ |
| (1j) | $\frac{dy}{dx} = \sec^3(x) + \sec(x)\tan^2(x)$ |
| (1k) | $\frac{dy}{dx} = e^x \cos(e^x)$ |
| (1l) | $\frac{dy}{dx} = \cot(x)$ |
| (1m) | $\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}$ |
| (1n) | $\frac{dy}{dx} = \frac{(x^2 + 1)\cos(x) - 2x \sin(x)}{(x^2 + 1)^2}$ |
| (1o) | $\frac{dy}{dx} = \frac{2x}{1 + x^4}$ |
| (1p) | $\frac{dy}{dx} = \frac{2x}{x^2 + 1}$ |
| (2) | $g'(4) = \frac{\sqrt{2}}{8}$ |
| (3a) | $y = 6x - 9$ |
| (3b) | $y = x - 1$ |
| (3c) | $y = x$ |
| (3d) | Ambiguous — likely typo |

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If you can provide the exact text of problem (3d), I can give the correct answer.

Let me know if you'd like explanations for any particular step!