Yes, for a single variableLet x, y be two given functionsConsiderf=(x+y)2.By the chain rule, we havef′=2(x+y)(x′+y′)=2(xx′+xy′+yx′+yy′)Now, expand f to obtainf=x2+2xy+y2Thus, we havef′=2xx′+2(xy)′+2yy′=2(xx′+(xy)′+yy′))It follows immediately that(xy)′=xy′+yx′Thus, we see that equality holds
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