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chain_rules

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Chain rules

\begin{eqnarray*} y & = & f(t) \\ t & = & g(x) \\ y & =& f(g(x)) \\ \frac {dy}{dx} & = & \frac {dy}{dt} * \frac {dt}{dx} \\ & & \frac {dy}{dt} = f'(t) = f'(g(x)) \\ & & \because{ \frac {dt}{dx} = g'(x) } \\ & & \frac {dy}{dx} = f'(g(x)) * g'(x) \\ \end{eqnarray*}

\begin{eqnarray*} y & = & (2x^2 + 1)^2 \\ t & = & 2x^2 + 1 \\ y & = & t^2 \\b t & = & 2x^2 + 1 \\ \frac{dy}{dt} & = & 2t \\ & = & 2 (2x^2 + 1) \\ & = & (4x^2 + 2) \\ \frac{dt}{dx} & = & 4x \\ \therefore{} \frac{dy}{dx} & = & \frac{dy}{dt}*\frac{dt}{dx} \\ & = & (4x^2 + 2) * 4x \\ & = & 16x^3 + 8x \\ \end{eqnarray*}

e.g.

y.hat = a + b * x
a = intercept
residuals = (y - y.hat)
d.sum.of.residuals^2 / d.intercept
= d.sum.of.residuals^2 / d.sum.of.residuals * d.sum.of.residuals / d.intercept
= (2 * residual) * d(y - y.hat)/d.intercept
= (2 * residual) * d(y - (a + bx))
= (2 * residual) * d(y - a - bx)
= (2 * residual) * -1
= -2 * residual

y.hat = a + b * x
b = slope
d.sum.of.square.res / d.slope
= d.sum.of.square.res / d.sum.of.res * d.sum.of.res / d.slope
= d.sum.of.square.res / d.slope
= (2 * residual) * d(y - a - bx)
= (2 * residual) * - x
= - 2 * x * residual

chain_rules.1754319575.txt.gz · Last modified: 2025/08/04 23:59 by hkimscil

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