Solve a problem

DerivativesCalculusCollegeabout 1:42

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

A narrated, step-by-step derivatives video lesson with captions. Final answer: f'(x) = (2x + 1)·ln(e^x + 1) / ((x² + x + 1)² + 1) + atan(x² + x + 1)·e^x / (e^x + 1).

SpeedLanguage

Chapters

Transcript

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.7 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Finding a derivative
$\frac{d}{dx}\left[\tan^{-1}\left({x}^{2}+ x+1\right) \ln\left({e}^{x}+1\right)\right]$
2. Steps, part 1
1. $\frac{d}{dx}\left[\tan^{-1}\left({x}^{2}+ x+1\right) \ln\left({e}^{x}+1\right)\right] = \frac{\left(2x+1\right) \ln\left({e}^{x}+1\right)}{{\left({x}^{2}+ x+1\right)}^{2}+1}+\frac{\tan^{-1}\left({x}^{2}+ x+1\right) {e}^{x}}{{e}^{x}+1}$
  Product rule: (uv)′ = u′v + uv′
2. $\frac{d}{dx}\left[\ln\left({e}^{x}+1\right)\right] = \frac{{e}^{x}}{{e}^{x}+1}$
  Chain rule — d/dx[ln u] = 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. Steps, part 2
1. $\frac{d}{dx}\left[\tan^{-1}\left({x}^{2}+ x+1\right)\right] = \frac{2x+1}{{\left({x}^{2}+ x+1\right)}^{2}+1}$
  Chain rule — d/dx[arctan u] = u′/(1 + u²)
2. $\frac{d}{dx}\left[{x}^{2}+ x+1\right] = 2x+1$
  Sum rule: differentiate each term separately
3. $\frac{d}{dx}\left[{x}^{2}+ x\right] = 2x+1$
  Sum rule: differentiate each term separately
4. $\frac{d}{dx}\left[{x}^{2}\right] = 2x$
  Power rule: d/dx[xⁿ] = n·xⁿ⁻¹
4. Steps, part 3
1. $f'(x) = \frac{\left(2x+1\right) \ln\left({e}^{x}+1\right)}{{\left({x}^{2}+ x+1\right)}^{2}+1}+\frac{\tan^{-1}\left({x}^{2}+ x+1\right) {e}^{x}}{{e}^{x}+1}$
  Combine and simplify
5. The answer
$f'(x) = \frac{\left(2x+1\right) \ln\left({e}^{x}+1\right)}{{\left({x}^{2}+ x+1\right)}^{2}+1}+\frac{\tan^{-1}\left({x}^{2}+ x+1\right) {e}^{x}}{{e}^{x}+1}$
f'(x) = (2x + 1)·ln(e^x + 1) / ((x² + x + 1)² + 1) + atan(x² + x + 1)·e^x / (e^x + 1)
6. Check
- Cross-checked numerically against an independent symbolic differentiator at 8 sample points ✓

Transcript

The problem. Here is the problem: derivative of (arctan(x squared plus x plus 1))(ln(e to the power x plus 1)). Find the derivative of (atan(x squared plus x plus 1)) (ln(e to the power x plus 1)) with respect to x.

Steps, part 1. Product rule: (uv) prime equals u prime v plus uv prime. Chain rule — d by d x[ln u] equals u prime /u. Sum rule: differentiate each term separately. Exponential rule: d by d x[eᵘ] equals eᵘ times u prime.

Steps, part 2. Chain rule — d by d x[arctan u] equals u prime /(1 plus u squared). Sum rule: differentiate each term separately. Sum rule: differentiate each term separately. Power rule: d by d x[x to the power n] equals n times x to the power n minus 1.

Steps, part 3. Combine and simplify.

The answer. So the answer is f prime of x equals (2x plus 1) times ln(e to the power x plus 1) / ((x squared plus x plus 1) squared plus 1) plus atan(x squared plus x plus 1) times e to the power x / (e to the power x plus 1). The derivative is built from the outside in: sums split into terms, constants factor out, products and quotients use their rules, and every composite function multiplies by the derivative of its inside (chain rule).

Check. Cross-checked numerically against an independent symbolic differentiator at 8 sample points, which checks out.

About derivatives

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

Derivative of (arctan(x^2+x+1))(ln(e^(x)+1)) (Lesson)