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4.8 Computing derivatives

F′⁢(t)=h⁢∂⁡f∂⁡x⁢(a+t⁢h,b+t⁢k)+k⁢∂⁡f∂⁡y⁢(a+t⁢h,b+t⁢k).F^{\prime}(t)=h{{\partial f}\over{\partial x}}(a+th,b+tk)+k{{\partial f}\over{% \partial y}}(a+th,b+tk).

Differentiating again, we obtain

F′′⁢(t)=dd⁢t⁢(h⁢∂⁡f∂⁡x⁢(a+t⁢h,b+t⁢k)+k⁢∂⁡f∂⁡y⁢(a+t⁢h,b+t⁢k))=F^{\prime\prime}(t)=\frac{d}{dt}\left(h{{\partial f}\over{\partial x}}(a+th,b+% tk)+k{{\partial f}\over{\partial y}}(a+th,b+tk)\right)=\hskip{28.452756pt}
h⁢dd⁢t⁢∂⁡f∂⁡x+k⁢dd⁢t⁢∂⁡f∂⁡y=h⁢(h⁢∂2⁡f∂⁡x2+k⁢∂2⁡f∂⁡y⁢∂⁡x)+k⁢(h⁢∂2⁡f∂⁡x⁢∂⁡y+k⁢∂2⁡f∂⁡y2)\hskip{14.226378pt}h{\frac{{\hbox{d}}}{{\hbox{d}}t}}{\frac{\partial f}{% \partial x}}+k{\frac{{\hbox{d}}}{{\hbox{d}}t}}{\frac{\partial f}{\partial y}}=% h\Bigl(h{{\partial^{2}f}\over{\partial x^{2}}}+k{{\partial^{2}f}\over{{% \partial y}{\partial x}}}\Bigr)+k\Bigl(h{{\partial^{2}f}\over{{\partial x}{% \partial y}}}+k{{\partial^{2}f}\over{\partial y^{2}}}\Bigr)
=h2⁢∂2⁡f∂⁡x2+2⁢h⁢k⁢∂2⁡f∂⁡x⁢∂⁡y+k2⁢∂2⁡f∂⁡y2,=h^{2}{{\partial^{2}f}\over{\partial x^{2}}}+2hk{{\partial^{2}f}\over{{% \partial x}{\partial y}}}+k^{2}{{\partial^{2}f}\over{\partial y^{2}}},

where the last step follows by Theorem 2.14 on the equality of mixed partial derivatives. On substituting these expressions for the derivatives of FF\, into the Maclaurin series, we obtain the stated formula for f⁢(a+h,b+k)f(a+h,b+k)\,.