Derivative of (sqrt(x^3))((e^(x)+1)^3): animated lesson
A narrated, step-by-step derivatives video lesson with captions. Final answer: f'(x) = (x²(e^x + 1)³·3/2 + 3sqrt(x³)²(e^x + 1)²e^x) / sqrt(x³).
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 (0.6 MB) · Captions (WebVTT)
Steps on screen
- 1. The problemFinding a derivative
- 2. Steps, part 1
- Product rule: (uv)′ = u′v + uv′
- Power rule with the chain rule: d/dx[uⁿ] = n·uⁿ⁻¹·u′
- Sum rule: differentiate each term separately
- Exponential rule: d/dx[eᵘ] = eᵘ·u′
- 3. Steps, part 2
- Chain rule — d/dx[√u] = u′/(2√u)
- Power rule: d/dx[xⁿ] = n·xⁿ⁻¹
- Combine and simplify
- 4. The answer
f'(x) = (x²(e^x + 1)³·3/2 + 3sqrt(x³)²(e^x + 1)²e^x) / sqrt(x³)
- 5. Check
- Cross-checked numerically against an independent symbolic differentiator at 8 sample points ✓
Transcript
The problem. Here is the problem: derivative of (sqrt(x cubed))((e to the power x plus 1) cubed). Find the derivative of (sqrt(x cubed)) ((e to the power x plus 1) cubed) with respect to x.
Steps, part 1. Product rule: (uv) prime equals u prime v plus uv prime. Power rule with the chain rule: d by d x[u to the power n] equals n times u to the power n minus 1 times u prime. 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[the square root of u] equals u prime /(2the square root of u. Power rule: d by d x[x to the power n] equals n times x to the power n minus 1. Combine and simplify.
The answer. So the answer is f prime of x equals (x squared(e to the power x plus 1) cubed times 3 over 2 plus 3sqrt(x cubed) squared(e to the power x plus 1) squarede to the power x / sqrt(x cubed). 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.
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