Home page for accesible maths 2 Chapter 2 contents

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

2.6 Differentiating using calculus rules

Example.

Find ∂⁡f∂⁡x{{\partial f}\over{\partial x}} and ∂⁡f∂⁡y{{\partial f}\over{\partial y}} for f⁢(x,y)=ex⁢y+x+yf(x,y)=e^{xy}+x+y.

Solution. In this case it may be helpful, when thinking of yy as a constant, to fix a value bb and consider the function ex⁢ye^{xy} as eb⁢xe^{bx}, which we think of as a function of xx only. Differentiating with respect to xx, we get:

∂⁡f∂⁡x=b⁢eb⁢x+1=y⁢ex⁢y+1.\frac{\partial f}{\partial x}=\,{be^{bx}+1}\,{=ye^{xy}+1.}

Similarly (or by symmetry), ∂⁡f∂⁡y=x⁢ex⁢y+1.\frac{\partial f}{\partial y}={xe^{xy}+1.}