Derivative of (ln(3x-2))/(cos(e^(x)+1)): animated lesson

1. 1. The problem
   Finding a derivative

   $\frac{d}{dx}\left[\frac{\ln\left(3x-2\right)}{\cos\left({e}^{x}+1\right)}\right]$
2. 2. Steps, part 1
   1. $$\frac{d}{dx}\left[\frac{\ln\left(3x-2\right)}{\cos\left({e}^{x}+1\right)}\right] = \frac{\frac{3}{3x-2}+\frac{\ln\left(3x-2\right) \sin\left({e}^{x}+1\right) {e}^{x}}{\cos\left({e}^{x}+1\right)}}{\cos\left({e}^{x}+1\right)}$$
      Quotient rule: (u/v)′ = (u′v − uv′)/v²
   2. $$\frac{d}{dx}\left[\cos\left({e}^{x}+1\right)\right] = -\left(\sin\left({e}^{x}+1\right) {e}^{x}\right)$$
      Chain rule — d/dx[cos u] = −sin u · u′
   3. $$\frac{d}{dx}\left[{e}^{x}+1\right] = {e}^{x}$$
      Sum rule: differentiate each term separately
   4. $$\frac{d}{dx}\left[{e}^{x}\right] = {e}^{x}$$
      Exponential rule: d/dx[eᵘ] = eᵘ·u′
3. 3. Steps, part 2
   1. $$\frac{d}{dx}\left[\ln\left(3x-2\right)\right] = \frac{3}{3x-2}$$
      Chain rule — d/dx[ln u] = u′/u
   2. $$\frac{d}{dx}\left[3x-2\right] = 3$$
      Difference rule: differentiate each term separately
   3. $$\frac{d}{dx}\left[3x\right] = 3$$
      Constant multiple rule
   4. $$f'(x) = \frac{\frac{3}{3x-2}+\frac{\ln\left(3x-2\right) \sin\left({e}^{x}+1\right) {e}^{x}}{\cos\left({e}^{x}+1\right)}}{\cos\left({e}^{x}+1\right)}$$
      Combine and simplify
4. 4. The answer
   $$f'(x) = \frac{\frac{3}{3x-2}+\frac{\ln\left(3x-2\right) \sin\left({e}^{x}+1\right) {e}^{x}}{\cos\left({e}^{x}+1\right)}}{\cos\left({e}^{x}+1\right)}$$

   f'(x) = (3 / (3 x - 2) + ln(3 x - 2)·sin(e^x + 1)·e^x / cos(e^x + 1)) / cos(e^x + 1)

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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