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Example 4

We let the scalar \alpha be defined by

\alpha = \boldsymbol{y}^T\boldsymbol{x},

where both \boldsymbol{y} and \boldsymbol{x} have the same length n , or if we wish to think of them as column vectors, they have dimensions n\times 1 . We assume that both \boldsymbol{y} and \boldsymbol{x} depend on a vector \boldsymbol{z} of the same length. To calculate the derivative of \alpha with respect to a given component z_k we need first to write out the inner product that defines \alpha as

\alpha = \sum_{i=0}^{n-1}y_ix_i,

and the partial derivative

\frac{\partial \alpha}{\partial z_k} = \sum_{i=0}^{n-1}\left(x_i\frac{\partial y_i}{\partial z_k}+y_i\frac{\partial x_i}{\partial z_k}\right),

for \forall k =0,1,2,\dots,n-1 . We can rewrite the partial derivative in a more compact form as

\frac{\partial \alpha}{\partial \boldsymbol{z}} = \boldsymbol{x}^T\frac{\partial \boldsymbol{y}}{\partial \boldsymbol{z}}+\boldsymbol{y}^T\frac{\partial \boldsymbol{x}}{\partial \boldsymbol{z}},

and if \boldsymbol{y}=\boldsymbol{x} we have

\frac{\partial \alpha}{\partial \boldsymbol{z}} = 2\boldsymbol{x}^T\frac{\partial \boldsymbol{x}}{\partial \boldsymbol{z}}.