finding the partial derivative with respect to x:
f(x,y)=2x+3y+xy-sin(x)+y^2
using the definition of a derivative:
lim(h-0) (f(x+h,y)-f(x,y))/h

Question

finding the partial derivative with respect to x:
f(x,y)=2x+3y+xy-sin(x)+y^2
using the definition of a derivative:
lim(h->0) (f(x+h,y)-f(x,y))/h

anonymous · Answer

$$f \left( x,y \right)=2x+3y+xy-\sin x+y ^{2}$$differentiate with respect to x, $$f _{x}=\lim_{h \rightarrow 0} \frac{2(x+h)+3y+(x+h)y-\sin(x+h)+y ^{2}-(2x+3y+xy-\sin(x)+y ^{2})}{h}$$ $$f _{x}=\lim_{h \rightarrow 0} \frac{2x+2h+3y+xy+hy-(\sin(x)\cos(h)+\cos(x)\sin(h))+y ^{2}-(2x+3y+xy-\sin(x)+y ^{2})}{h}$$simplifying,$$f _{x}=\lim_{h \rightarrow 0} \frac{2h+2y-[\sin(x)cox(h)-1]-\cos(x)\sin(h)}{h}$$
$$f _{x}=\lim_{h \rightarrow 0} \frac{h(2+y)}{h}-\frac{\sin(x)\cos(h)-1}{h}-\frac{\cos(x)\sin(h)}{h}$$since $$\lim_{h \rightarrow 0}\frac{\cos(h)-1}{h}=0, and \lim_{h \rightarrow 0}\frac{\sin(h)}{h}=1,$$then $$f _{x}=\lim_{h \rightarrow 0} 2+y-\cos(x)$$ $$f _{x}=2+y-\cos(x)$$