Derivative of vector w.r.t. scalar












0














I am struggling with backpropagating BatchNormalization. For that, I would like to take the partial derivative of a vector valued function with respect to a scalar. The simplified function looks like this.



$$
overrightarrow{f}(overrightarrow{x}, y) = overrightarrow{x} + (y,y,y) =
begin{bmatrix}
x_{1} + y
\
x_{2} + y
\
x_{3} + y
end{bmatrix}
$$



I can see that
$$frac{partial{f_i}}{partial{y}} = 1$$
And following this post the partial derivative for the vector-valued function should equal



$$frac{partial{overrightarrow{f}}}{partial{y}} = begin{bmatrix}
1
\
1
\
1
end{bmatrix}$$



But from my knowledge of backprop I thought that the derivative should have the same shape as the variable which is obviously not the case because $dim(frac{partial{overrightarrow{f}}}{partial{y}}) neq dim(y)$



But let's say we have a second function $g(overrightarrow{f}) = g(f_{1}(x_{1}, y), cdots)$, then with the multivariable variable chain rule we would obtain:
$$
frac{partial{g}}{partial{y}} = sum_{i=1}^{3}frac{partial{f_{i}}}{partial{y}} = 3
$$



I would like to ask if my intuition is correct.
Thanks in advance.






calculus multivariable-calculus







share|cite|improve this question





























    0














    I am struggling with backpropagating BatchNormalization. For that, I would like to take the partial derivative of a vector valued function with respect to a scalar. The simplified function looks like this.



    $$
    overrightarrow{f}(overrightarrow{x}, y) = overrightarrow{x} + (y,y,y) =
    begin{bmatrix}
    x_{1} + y
    \
    x_{2} + y
    \
    x_{3} + y
    end{bmatrix}
    $$



    I can see that
    $$frac{partial{f_i}}{partial{y}} = 1$$
    And following this post the partial derivative for the vector-valued function should equal



    $$frac{partial{overrightarrow{f}}}{partial{y}} = begin{bmatrix}
    1
    \
    1
    \
    1
    end{bmatrix}$$



    But from my knowledge of backprop I thought that the derivative should have the same shape as the variable which is obviously not the case because $dim(frac{partial{overrightarrow{f}}}{partial{y}}) neq dim(y)$



    But let's say we have a second function $g(overrightarrow{f}) = g(f_{1}(x_{1}, y), cdots)$, then with the multivariable variable chain rule we would obtain:
    $$
    frac{partial{g}}{partial{y}} = sum_{i=1}^{3}frac{partial{f_{i}}}{partial{y}} = 3
    $$



    I would like to ask if my intuition is correct.
    Thanks in advance.






    calculus multivariable-calculus







    share|cite|improve this question



























      0












      0








      0







      I am struggling with backpropagating BatchNormalization. For that, I would like to take the partial derivative of a vector valued function with respect to a scalar. The simplified function looks like this.



      $$
      overrightarrow{f}(overrightarrow{x}, y) = overrightarrow{x} + (y,y,y) =
      begin{bmatrix}
      x_{1} + y
      \
      x_{2} + y
      \
      x_{3} + y
      end{bmatrix}
      $$



      I can see that
      $$frac{partial{f_i}}{partial{y}} = 1$$
      And following this post the partial derivative for the vector-valued function should equal



      $$frac{partial{overrightarrow{f}}}{partial{y}} = begin{bmatrix}
      1
      \
      1
      \
      1
      end{bmatrix}$$



      But from my knowledge of backprop I thought that the derivative should have the same shape as the variable which is obviously not the case because $dim(frac{partial{overrightarrow{f}}}{partial{y}}) neq dim(y)$



      But let's say we have a second function $g(overrightarrow{f}) = g(f_{1}(x_{1}, y), cdots)$, then with the multivariable variable chain rule we would obtain:
      $$
      frac{partial{g}}{partial{y}} = sum_{i=1}^{3}frac{partial{f_{i}}}{partial{y}} = 3
      $$



      I would like to ask if my intuition is correct.
      Thanks in advance.






      calculus multivariable-calculus







      share|cite|improve this question















      I am struggling with backpropagating BatchNormalization. For that, I would like to take the partial derivative of a vector valued function with respect to a scalar. The simplified function looks like this.



      $$
      overrightarrow{f}(overrightarrow{x}, y) = overrightarrow{x} + (y,y,y) =
      begin{bmatrix}
      x_{1} + y
      \
      x_{2} + y
      \
      x_{3} + y
      end{bmatrix}
      $$



      I can see that
      $$frac{partial{f_i}}{partial{y}} = 1$$
      And following this post the partial derivative for the vector-valued function should equal



      $$frac{partial{overrightarrow{f}}}{partial{y}} = begin{bmatrix}
      1
      \
      1
      \
      1
      end{bmatrix}$$



      But from my knowledge of backprop I thought that the derivative should have the same shape as the variable which is obviously not the case because $dim(frac{partial{overrightarrow{f}}}{partial{y}}) neq dim(y)$



      But let's say we have a second function $g(overrightarrow{f}) = g(f_{1}(x_{1}, y), cdots)$, then with the multivariable variable chain rule we would obtain:
      $$
      frac{partial{g}}{partial{y}} = sum_{i=1}^{3}frac{partial{f_{i}}}{partial{y}} = 3
      $$



      I would like to ask if my intuition is correct.
      Thanks in advance.






      calculus multivariable-calculus




      calculus multivariable-calculus






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      share|cite|improve this question













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      edited Jan 3 at 23:12

























      asked Jan 3 at 22:08









      oezguensi

      133




      133






















          2 Answers
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          0














          Comment: you cannot sum a vector $x(3 times 1)$ and a scalar $y$.






          share|cite|improve this answer





















          • You CAN, it just doesn't mean anything on its own ($textbf{x}+y=textbf{x}+y$). In any case, I think the OP meant $textbf{x}+(y,y,y)$.
            – R. Burton
            Jan 3 at 22:27












          • In this case, define $y=(y_1,y_2,y_3)$, compute the $3 times 3$ matrix of partial derivatives $$ frac{partial f^T}{partial y}(x,y) $$ and evaluate it at $y=(1,1,1)$, which gives the $3 times 3$ identity matrix.
            – Bertrand
            Jan 3 at 22:37












          • Sorry, I meant and element wise addition such as $mathbf{x} + (y,y,y)$
            – oezguensi
            Jan 3 at 23:14



















          0














          I assume you mean
          $$textbf{f}(textbf{x},y)=textbf{x}+(y,y,y)=(x_1+y,x_2+y,x_3+y)$$
          In which case
          $$frac{partialtextbf{f}}{partial y}=left(frac{partial f_1}{partial y},frac{partial f_2}{partial y},frac{partial f_3}{partial y}right)=(1,1,1)$$
          The dimensions of $y$ are not particularly relevant, and in general the dimensions of the variable you take the derivative with respect to does not matter. In general, for any scalar $x$ and vector valued function $textbf{f}$, $dimleft(frac{partial textbf{f}}{partial x}right)=dim(textbf{f})neq dim(x)$.






          share|cite|improve this answer





















          • Sorry, I meant an element wise addition. Thank you for the answer.
            – oezguensi
            Jan 3 at 23:13











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0














          Comment: you cannot sum a vector $x(3 times 1)$ and a scalar $y$.






          share|cite|improve this answer





















          • You CAN, it just doesn't mean anything on its own ($textbf{x}+y=textbf{x}+y$). In any case, I think the OP meant $textbf{x}+(y,y,y)$.
            – R. Burton
            Jan 3 at 22:27












          • In this case, define $y=(y_1,y_2,y_3)$, compute the $3 times 3$ matrix of partial derivatives $$ frac{partial f^T}{partial y}(x,y) $$ and evaluate it at $y=(1,1,1)$, which gives the $3 times 3$ identity matrix.
            – Bertrand
            Jan 3 at 22:37












          • Sorry, I meant and element wise addition such as $mathbf{x} + (y,y,y)$
            – oezguensi
            Jan 3 at 23:14
















          0














          Comment: you cannot sum a vector $x(3 times 1)$ and a scalar $y$.






          share|cite|improve this answer





















          • You CAN, it just doesn't mean anything on its own ($textbf{x}+y=textbf{x}+y$). In any case, I think the OP meant $textbf{x}+(y,y,y)$.
            – R. Burton
            Jan 3 at 22:27












          • In this case, define $y=(y_1,y_2,y_3)$, compute the $3 times 3$ matrix of partial derivatives $$ frac{partial f^T}{partial y}(x,y) $$ and evaluate it at $y=(1,1,1)$, which gives the $3 times 3$ identity matrix.
            – Bertrand
            Jan 3 at 22:37












          • Sorry, I meant and element wise addition such as $mathbf{x} + (y,y,y)$
            – oezguensi
            Jan 3 at 23:14














          0












          0








          0






          Comment: you cannot sum a vector $x(3 times 1)$ and a scalar $y$.






          share|cite|improve this answer












          Comment: you cannot sum a vector $x(3 times 1)$ and a scalar $y$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 22:24









          Bertrand

          413




          413












          • You CAN, it just doesn't mean anything on its own ($textbf{x}+y=textbf{x}+y$). In any case, I think the OP meant $textbf{x}+(y,y,y)$.
            – R. Burton
            Jan 3 at 22:27












          • In this case, define $y=(y_1,y_2,y_3)$, compute the $3 times 3$ matrix of partial derivatives $$ frac{partial f^T}{partial y}(x,y) $$ and evaluate it at $y=(1,1,1)$, which gives the $3 times 3$ identity matrix.
            – Bertrand
            Jan 3 at 22:37












          • Sorry, I meant and element wise addition such as $mathbf{x} + (y,y,y)$
            – oezguensi
            Jan 3 at 23:14


















          • You CAN, it just doesn't mean anything on its own ($textbf{x}+y=textbf{x}+y$). In any case, I think the OP meant $textbf{x}+(y,y,y)$.
            – R. Burton
            Jan 3 at 22:27












          • In this case, define $y=(y_1,y_2,y_3)$, compute the $3 times 3$ matrix of partial derivatives $$ frac{partial f^T}{partial y}(x,y) $$ and evaluate it at $y=(1,1,1)$, which gives the $3 times 3$ identity matrix.
            – Bertrand
            Jan 3 at 22:37












          • Sorry, I meant and element wise addition such as $mathbf{x} + (y,y,y)$
            – oezguensi
            Jan 3 at 23:14
















          You CAN, it just doesn't mean anything on its own ($textbf{x}+y=textbf{x}+y$). In any case, I think the OP meant $textbf{x}+(y,y,y)$.
          – R. Burton
          Jan 3 at 22:27






          You CAN, it just doesn't mean anything on its own ($textbf{x}+y=textbf{x}+y$). In any case, I think the OP meant $textbf{x}+(y,y,y)$.
          – R. Burton
          Jan 3 at 22:27














          In this case, define $y=(y_1,y_2,y_3)$, compute the $3 times 3$ matrix of partial derivatives $$ frac{partial f^T}{partial y}(x,y) $$ and evaluate it at $y=(1,1,1)$, which gives the $3 times 3$ identity matrix.
          – Bertrand
          Jan 3 at 22:37






          In this case, define $y=(y_1,y_2,y_3)$, compute the $3 times 3$ matrix of partial derivatives $$ frac{partial f^T}{partial y}(x,y) $$ and evaluate it at $y=(1,1,1)$, which gives the $3 times 3$ identity matrix.
          – Bertrand
          Jan 3 at 22:37














          Sorry, I meant and element wise addition such as $mathbf{x} + (y,y,y)$
          – oezguensi
          Jan 3 at 23:14




          Sorry, I meant and element wise addition such as $mathbf{x} + (y,y,y)$
          – oezguensi
          Jan 3 at 23:14











          0














          I assume you mean
          $$textbf{f}(textbf{x},y)=textbf{x}+(y,y,y)=(x_1+y,x_2+y,x_3+y)$$
          In which case
          $$frac{partialtextbf{f}}{partial y}=left(frac{partial f_1}{partial y},frac{partial f_2}{partial y},frac{partial f_3}{partial y}right)=(1,1,1)$$
          The dimensions of $y$ are not particularly relevant, and in general the dimensions of the variable you take the derivative with respect to does not matter. In general, for any scalar $x$ and vector valued function $textbf{f}$, $dimleft(frac{partial textbf{f}}{partial x}right)=dim(textbf{f})neq dim(x)$.






          share|cite|improve this answer





















          • Sorry, I meant an element wise addition. Thank you for the answer.
            – oezguensi
            Jan 3 at 23:13
















          0














          I assume you mean
          $$textbf{f}(textbf{x},y)=textbf{x}+(y,y,y)=(x_1+y,x_2+y,x_3+y)$$
          In which case
          $$frac{partialtextbf{f}}{partial y}=left(frac{partial f_1}{partial y},frac{partial f_2}{partial y},frac{partial f_3}{partial y}right)=(1,1,1)$$
          The dimensions of $y$ are not particularly relevant, and in general the dimensions of the variable you take the derivative with respect to does not matter. In general, for any scalar $x$ and vector valued function $textbf{f}$, $dimleft(frac{partial textbf{f}}{partial x}right)=dim(textbf{f})neq dim(x)$.






          share|cite|improve this answer





















          • Sorry, I meant an element wise addition. Thank you for the answer.
            – oezguensi
            Jan 3 at 23:13














          0












          0








          0






          I assume you mean
          $$textbf{f}(textbf{x},y)=textbf{x}+(y,y,y)=(x_1+y,x_2+y,x_3+y)$$
          In which case
          $$frac{partialtextbf{f}}{partial y}=left(frac{partial f_1}{partial y},frac{partial f_2}{partial y},frac{partial f_3}{partial y}right)=(1,1,1)$$
          The dimensions of $y$ are not particularly relevant, and in general the dimensions of the variable you take the derivative with respect to does not matter. In general, for any scalar $x$ and vector valued function $textbf{f}$, $dimleft(frac{partial textbf{f}}{partial x}right)=dim(textbf{f})neq dim(x)$.






          share|cite|improve this answer












          I assume you mean
          $$textbf{f}(textbf{x},y)=textbf{x}+(y,y,y)=(x_1+y,x_2+y,x_3+y)$$
          In which case
          $$frac{partialtextbf{f}}{partial y}=left(frac{partial f_1}{partial y},frac{partial f_2}{partial y},frac{partial f_3}{partial y}right)=(1,1,1)$$
          The dimensions of $y$ are not particularly relevant, and in general the dimensions of the variable you take the derivative with respect to does not matter. In general, for any scalar $x$ and vector valued function $textbf{f}$, $dimleft(frac{partial textbf{f}}{partial x}right)=dim(textbf{f})neq dim(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 22:40









          R. Burton

          34919




          34919












          • Sorry, I meant an element wise addition. Thank you for the answer.
            – oezguensi
            Jan 3 at 23:13


















          • Sorry, I meant an element wise addition. Thank you for the answer.
            – oezguensi
            Jan 3 at 23:13
















          Sorry, I meant an element wise addition. Thank you for the answer.
          – oezguensi
          Jan 3 at 23:13




          Sorry, I meant an element wise addition. Thank you for the answer.
          – oezguensi
          Jan 3 at 23:13


















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