Computing a derivative (bonus)

Optional bonus: differentiation rules, partial derivatives and the exact slope of the crêpe and admission errors

12 exercises

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  1. Lesson: slope of a formula

    In level 2, your zoom machine found the best crêpe line by trying finer and finer settings. With two settings, it is fast. With ten thousand, trying is no longer enough: you need to know which way to move each…

  2. Exact slope

    The on-board calculator of the crêpe stand tracks five curves, all functions of the same setting xx: - sales according to temperature: f1(x)=2x3−5x2+x−7f_1(x) = 2x^3 - 5x^2 + x - 7; - attendance after opening: f2(x)=x e−xf_2(x) = x\,e^{-x}; -…

  3. Slope of the squasher

    In level 2, the admissions table passed each score zz through the function you invented to squash any number between 0 and 1: σ(z)=11+e−z\sigma(z) = \dfrac{1}{1 + e^{-z}}. The table's sensor only transmits the output…

  4. Wrong step

    A student differentiates f(x)=x2e−xf(x) = x^2 e^{-x}: 1. f=u⋅vf = u \cdot v with u=x2u = x^2 and v=e−xv = e^{-x}; 2. u′=2xu' = 2x; 3. v′=e−xv' = e^{-x}; 4. f′(x)=u′v+uv′=2x e−x+x2e−xf'(x) = u'v + uv' = 2x\,e^{-x} + x^2 e^{-x}. Which step is wrong?

  5. Second wrong step

    A student differentiates f(x)=ln⁡(1+x2)f(x) = \ln(1 + x^2), then computes the slope at x=1x = 1: 1. f=ln⁡uf = \ln u with u=1+x2u = 1 + x^2; 2. (ln⁡u)′=1u(\ln u)' = \frac{1}{u}; 3. so f′(x)=11+x2f'(x) = \frac{1}{1 + x^2}; 4. f′(1)=12f'(1) = \frac{1}{2}. Which step…

  6. Point on the line

    To tune the line of the lying points, we compute the slope of the mean absolute error with respect to the setting. For each point, the slope of ∣r∣|r| (where rr is the gap between the point and the line) is +1+1 if…

  7. Lesson: several settings

    The error of the crêpe line depends on two settings, aa and bb. To know how it reacts to aa alone, we freeze bb: it becomes a constant, and we differentiate with respect to aa with the usual rules. The…

  8. Two settings

    The stand's dashboard shows three indicators that depend on two settings aa and bb: - f1(a,b)=a3b−2ab2+5bf_1(a, b) = a^3 b - 2ab^2 + 5b; - f2(a,b)=eab−af_2(a, b) = e^{ab} - a; - f3(a,b)=(14a+b−190)2f_3(a, b) = (14a + b - 190)^2, the squared gap of the line…

  9. Slope of the error

    Go back to the crêpe line y=ax+by = ax + b and to the lying points. To tune the line, the calculator needs the exact slope of the error with respect to each of the two settings, over nn records (xi,yi)(x_i, y_i): -…

  10. Slope of admission

    The school's admission board predicts whether a student is admitted from their grade xx with p=σ(wx+b)p = \sigma(w x + b), where σ(z)=11+e−z\sigma(z) = \frac{1}{1 + e^{-z}}. Over nn files (xi,yi)(x_i, y_i), where yiy_i is 1 if the student…

  11. Lesson: what you can differentiate

    You have differentiated whole losses. With the gap ri=y^i−yir_i = \hat{y}_i - y_i of each record: - the slope of the mean of the squares of a line is 2n∑iri (xi,1)\frac{2}{n}\sum_i r_i\,(x_i, 1); - that of the mean of the absolute gaps…

  12. Two thousand sensors

    The crêpe line has grown: the calculator now predicts y^=Xw\hat{y} = Xw from dd sensors, over nn records. XX has nn rows (the records) and dd columns (the sensors), ww holds the dd settings, and the error is the…