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2.7 Differentiating using calculus rules

Example.

To find ∂⁡f∂⁡x{{\partial f}\over{\partial x}} and ∂⁡f∂⁡y{{\partial f}\over{\partial y}} for f⁢(x,y)=(x2+x⁢y)⁢sin⁡(y2+x⁢y).f(x,y)=(x^{2}+xy)\sin(y^{2}+xy).

Solution. To calculate ∂⁡f∂⁡x\frac{\partial f}{\partial x}, think of yy as constant, say y=by=b. Then f⁢(x,b)=(x2+b⁢x)⁢sin⁡(b⁢x+b2)f(x,b)=(x^{2}+bx)\sin(bx+b^{2}). So, by the product rule:

∂⁡f∂⁡x=(2⁢x+b)⁢sin⁡(b⁢x+b2)+b⁢(x2+b⁢x)⁢cos⁡(b⁢x+b2)\frac{\partial f}{\partial x}=\,{(2x+b)\sin(bx+b^{2})+b(x^{2}+bx)\cos(bx+b^{2})}
=(2⁢x+y)⁢sin⁡(x⁢y+y2)+y⁢(x2+x⁢y)⁢cos⁡(x⁢y+y2).{=(2x+y)\sin(xy+y^{2})+y(x^{2}+xy)\cos(xy+y^{2}).}

Similarly, setting x=ax=a: ∂⁡f∂⁡y=(a⁢y+a2)′⁢sin⁡(y2+a⁢y)+(a⁢y+a2)⁢(sin⁡(y2+a⁢y))′\frac{\partial f}{\partial y}=\,{(ay+a^{2})^{\prime}\sin(y^{2}+ay)+(ay+a^{2})(% \sin(y^{2}+ay))^{\prime}}

=x⁢sin⁡(x2+x⁢y)+(x+2⁢y)⁢(x2+x⁢y)⁢cos⁡(x2+x⁢y).={x\sin(x^{2}+xy)+(x+2y)(x^{2}+xy)\cos(x^{2}+xy).}