Derivative of (e^(e^(x)+1))(cos(x^2+1)): animated lesson

1. 1. The problem
   Finding a derivative

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

   f'(x) = e^(e^x + 1 + x)·cos(x² + 1) - 2e^(e^x + 1)·sin(x² + 1)·x

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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