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Derivation on equation with summation symbol


Derivatives - Show equalityHow do you read a partial differential equation?How many solutions for an equation with simple restrictionsCompute derivative with respect to a matrixPartial differentiating implicitlyDerivative of a vector by a matrixDerivation of derivative of multivariate Gaussian w.r.t. covariance matrixHow to derive the substituted Partial Differential EquationHow to anti-derivative the partial derivative of a multivariate function across all variables to end up with the original functionPartial derivative of a function within a function, lambda calculus













0












$begingroup$


I have an equation where:



$J_t = frac12sum(haty-y)^2$ where $haty$ is a vector of $hat y_1,hat y_2,...hat y_k$.



If i were to differentiate $J_t$ in terms of $y_1$, is it right to assume that the $sum$ symbol doesn't affect the differentiation process?



Hence,



$fracpartial J_tpartial y_1 = (hat y_1 - y)$



I would like to know if the above is a correct way to differentiate?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Mar 30 at 4:11
















0












$begingroup$


I have an equation where:



$J_t = frac12sum(haty-y)^2$ where $haty$ is a vector of $hat y_1,hat y_2,...hat y_k$.



If i were to differentiate $J_t$ in terms of $y_1$, is it right to assume that the $sum$ symbol doesn't affect the differentiation process?



Hence,



$fracpartial J_tpartial y_1 = (hat y_1 - y)$



I would like to know if the above is a correct way to differentiate?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Mar 30 at 4:11














0












0








0





$begingroup$


I have an equation where:



$J_t = frac12sum(haty-y)^2$ where $haty$ is a vector of $hat y_1,hat y_2,...hat y_k$.



If i were to differentiate $J_t$ in terms of $y_1$, is it right to assume that the $sum$ symbol doesn't affect the differentiation process?



Hence,



$fracpartial J_tpartial y_1 = (hat y_1 - y)$



I would like to know if the above is a correct way to differentiate?










share|cite|improve this question









$endgroup$




I have an equation where:



$J_t = frac12sum(haty-y)^2$ where $haty$ is a vector of $hat y_1,hat y_2,...hat y_k$.



If i were to differentiate $J_t$ in terms of $y_1$, is it right to assume that the $sum$ symbol doesn't affect the differentiation process?



Hence,



$fracpartial J_tpartial y_1 = (hat y_1 - y)$



I would like to know if the above is a correct way to differentiate?







discrete-mathematics partial-derivative






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 30 at 3:42









MaxxxMaxxx

1255




1255







  • 1




    $begingroup$
    Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Mar 30 at 4:11













  • 1




    $begingroup$
    Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Mar 30 at 4:11








1




1




$begingroup$
Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 4:11





$begingroup$
Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 4:11











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