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Question Number 73274 by peter frank last updated on 09/Nov/19
Commented by kaivan.ahmadi last updated on 09/Nov/19
t=0⇒x=1,y=0⇒∂f∂t=∂f∂x×∂x∂t+∂f∂y×∂y∂t=(siny+excosy)(2t)+(xcosy−exsiny)(2t)∣t=0=e×0+(1−0)(0)=0
Commented by peter frank last updated on 09/Nov/19
thankyou
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