Page of handwritten calculus notes covering derivative rules for exponential and logarithmic functions, with examples and practice problems.

Thank you for uploading the image. It contains a worksheet titled “Derivative of Exponential and Logarithmic Functions — Logarithmic Differentiation” with 12 problems asking to find the derivative of various functions.

I’ll solve each problem step-by-step, using appropriate differentiation rules (chain rule, product rule, quotient rule, logarithmic differentiation where needed), and explain the reasoning.

---

Useful Formulas (from the sheet):

- $\frac{d}{dx} e^x = e^x$
- $\frac{d}{dx} a^x = a^x \ln a$
- $\frac{d}{dx} \ln(x) = \frac{1}{x}$
- $\frac{d}{dx} \log_a(x) = \frac{1}{x \ln a}$
- $\frac{d}{dx} [f(x)]^{g(x)} = [f(x)]^{g(x)} \cdot \left[ g'(x)\ln f(x) + \frac{g(x)f'(x)}{f(x)} \right]$

---

## ✔ Problem 1: $ y = x^x $

This is a classic case for logarithmic differentiation.

Step 1: Take natural log of both sides:

$$
\ln y = \ln(x^x) = x \ln x
$$

Step 2: Differentiate implicitly:

$$
\frac{1}{y} \cdot y' = \frac{d}{dx}(x \ln x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1
$$

Step 3: Multiply both sides by $ y $:

$$
y' = y (\ln x + 1) = x^x (\ln x + 1)
$$

✔ Answer: $ \boxed{y' = x^x (\ln x + 1)} $

---

## ✔ Problem 2: $ y = e^{x^2} $

Use chain rule.

Let $ u = x^2 $, so $ y = e^u $

$$
\frac{dy}{dx} = e^u \cdot \frac{du}{dx} = e^{x^2} \cdot 2x
$$

✔ Answer: $ \boxed{y' = 2x e^{x^2}} $

---

## ✔ Problem 3: $ y = 5^{x^2} $

Recall: $ \frac{d}{dx} a^{u(x)} = a^{u(x)} \ln a \cdot u'(x) $

Here, $ a = 5 $, $ u(x) = x^2 $, so:

$$
y' = 5^{x^2} \ln 5 \cdot 2x
$$

✔ Answer: $ \boxed{y' = 2x \ln 5 \cdot 5^{x^2}} $

---

## ✔ Problem 4: $ y = e^{-\sin(4x)} $

Chain rule again.

Let $ u = -\sin(4x) $, so $ y = e^u $

$$
y' = e^u \cdot \frac{du}{dx} = e^{-\sin(4x)} \cdot (-\cos(4x) \cdot 4) = -4 \cos(4x) e^{-\sin(4x)}
$$

✔ Answer: $ \boxed{y' = -4 \cos(4x) e^{-\sin(4x)}} $

---

## ✔ Problem 5: $ y = \sqrt{1 + 2e^x} $

Write as $ y = (1 + 2e^x)^{1/2} $

Use chain rule:

$$
y' = \frac{1}{2}(1 + 2e^x)^{-1/2} \cdot (2e^x) = \frac{2e^x}{2\sqrt{1 + 2e^x}} = \frac{e^x}{\sqrt{1 + 2e^x}}
$$

✔ Answer: $ \boxed{y' = \frac{e^x}{\sqrt{1 + 2e^x}}} $

---

## ✔ Problem 6: $ y = \frac{x^2 + e^x}{x - e^x} $

Use quotient rule: $ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} $

Let:
- $ u = x^2 + e^x $ → $ u' = 2x + e^x $
- $ v = x - e^x $ → $ v' = 1 - e^x $

Then:

$$
y' = \frac{(2x + e^x)(x - e^x) - (x^2 + e^x)(1 - e^x)}{(x - e^x)^2}
$$

You can leave it like this, or expand numerator if required.

✔ Answer: $ \boxed{y' = \frac{(2x + e^x)(x - e^x) - (x^2 + e^x)(1 - e^x)}{(x - e^x)^2}} $

*(Note: Expanding is messy but doable if needed.)*

---

## ✔ Problem 7: $ y = \tan(e^{3x}) $

Chain rule: outer function is tan(u), inner is $ u = e^{3x} $

$$
y' = \sec^2(e^{3x}) \cdot \frac{d}{dx}(e^{3x}) = \sec^2(e^{3x}) \cdot 3e^{3x}
$$

✔ Answer: $ \boxed{y' = 3e^{3x} \sec^2(e^{3x})} $

---

## ✔ Problem 8: $ y = e^{e^x} $

Chain rule: let $ u = e^x $, so $ y = e^u $

$$
y' = e^u \cdot u' = e^{e^x} \cdot e^x = e^{x + e^x}
$$

✔ Answer: $ \boxed{y' = e^{x + e^x}} $

---

## ✔ Problem 9: $ y = \ln(x^2 \sqrt{x}) $

First, simplify using log rules:

$$
y = \ln(x^2) + \ln(\sqrt{x}) = 2\ln x + \frac{1}{2}\ln x = \frac{5}{2} \ln x
$$

Now differentiate:

$$
y' = \frac{5}{2} \cdot \frac{1}{x} = \frac{5}{2x}
$$

✔ Answer: $ \boxed{y' = \frac{5}{2x}} $

---

## ✔ Problem 10: $ y = \ln(\cos(5x)) $

Chain rule:

$$
y' = \frac{1}{\cos(5x)} \cdot (-\sin(5x)) \cdot 5 = -5 \tan(5x)
$$

✔ Answer: $ \boxed{y' = -5 \tan(5x)} $

---

## ✔ Problem 11: $ y = \log_3\left(\frac{2}{x^2 + 1}\right) $

Recall: $ \log_a(u) = \frac{\ln u}{\ln a} $, so

$$
y = \frac{1}{\ln 3} \ln\left( \frac{2}{x^2 + 1} \right) = \frac{1}{\ln 3} \left[ \ln 2 - \ln(x^2 + 1) \right]
$$

Differentiate:

$$
y' = \frac{1}{\ln 3} \left[ 0 - \frac{2x}{x^2 + 1} \right] = -\frac{2x}{(x^2 + 1)\ln 3}
$$

✔ Answer: $ \boxed{y' = -\frac{2x}{(x^2 + 1)\ln 3}} $

---

## ✔ Problem 12: $ y = \ln(\sqrt{x} \ln x) $

First, simplify inside the log:

$$
y = \ln\left( x^{1/2} \ln x \right) = \ln(x^{1/2}) + \ln(\ln x) = \frac{1}{2} \ln x + \ln(\ln x)
$$

Differentiate term by term:

- $ \frac{d}{dx} \left( \frac{1}{2} \ln x \right) = \frac{1}{2x} $
- $ \frac{d}{dx} \ln(\ln x) = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x} $

So,

$$
y' = \frac{1}{2x} + \frac{1}{x \ln x}
$$

You can factor out $ \frac{1}{x} $:

$$
y' = \frac{1}{x} \left( \frac{1}{2} + \frac{1}{\ln x} \right)
$$

✔ Answer: $ \boxed{y' = \frac{1}{2x} + \frac{1}{x \ln x}} $

---

## ✔ Final Summary of Answers:

| # | Function | Derivative |
|---|----------------------------------|----------------------------------------------------------------------------|
| 1 | $ y = x^x $ | $ x^x (\ln x + 1) $ |
| 2 | $ y = e^{x^2} $ | $ 2x e^{x^2} $ |
| 3 | $ y = 5^{x^2} $ | $ 2x \ln 5 \cdot 5^{x^2} $ |
| 4 | $ y = e^{-\sin(4x)} $ | $ -4 \cos(4x) e^{-\sin(4x)} $ |
| 5 | $ y = \sqrt{1 + 2e^x} $ | $ \frac{e^x}{\sqrt{1 + 2e^x}} $ |
| 6 | $ y = \frac{x^2 + e^x}{x - e^x} $ | $ \frac{(2x + e^x)(x - e^x) - (x^2 + e^x)(1 - e^x)}{(x - e^x)^2} $ |
| 7 | $ y = \tan(e^{3x}) $ | $ 3e^{3x} \sec^2(e^{3x}) $ |
| 8 | $ y = e^{e^x} $ | $ e^{x + e^x} $ |
| 9 | $ y = \ln(x^2 \sqrt{x}) $ | $ \frac{5}{2x} $ |
|10 | $ y = \ln(\cos(5x)) $ | $ -5 \tan(5x) $ |
|11 | $ y = \log_3\left(\frac{2}{x^2+1}\right) $ | $ -\frac{2x}{(x^2 + 1)\ln 3} $ |
|12 | $ y = \ln(\sqrt{x} \ln x) $ | $ \frac{1}{2x} + \frac{1}{x \ln x} $ |

---

Let me know if you want any of these expanded further or explained in more detail!