Derivative of (e^(2x))+(ln(3x-2)): animated lesson

1. 1. The problem
   Finding a derivative

   $\frac{d}{dx}\left[{e}^{2x}+\ln\left(3x-2\right)\right]$
2. 2. Steps, part 1
   1. $$\frac{d}{dx}\left[{e}^{2x}+\ln\left(3x-2\right)\right] = 2{e}^{2x}+\frac{3}{3x-2}$$
      Sum rule: differentiate each term separately
   2. $$\frac{d}{dx}\left[\ln\left(3x-2\right)\right] = \frac{3}{3x-2}$$
      Chain rule — d/dx[ln u] = u′/u
   3. $$\frac{d}{dx}\left[3x-2\right] = 3$$
      Difference rule: differentiate each term separately
   4. $$\frac{d}{dx}\left[3x\right] = 3$$
      Constant multiple rule
3. 3. Steps, part 2
   1. $$\frac{d}{dx}\left[{e}^{2x}\right] = 2{e}^{2x}$$
      Exponential rule: d/dx[eᵘ] = eᵘ·u′
   2. $$\frac{d}{dx}\left[2x\right] = 2$$
      Constant multiple rule
   3. $$f'(x) = 2{e}^{2x}+\frac{3}{3x-2}$$
      Combine and simplify
4. 4. The answer
   $$f'(x) = 2{e}^{2x}+\frac{3}{3x-2}$$

   f'(x) = 2e^(2x) + 3 / (3 x - 2)

5. 5. Check
   • ✓Cross-checked numerically against an independent symbolic differentiator at 8 sample points ✓

Differentiate with the power, product, quotient and chain rules, including higher derivatives.

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