(x-y)^2 = 1
are u from Sri Lanka?
ya
ah nice to meet you then so with what u need help
\[x-y= \pm1\]
can u pls gv me derivative to (x - y)^2 = 1
with respect to what?
to y
okay
\[(x-y)^{2}= x ^{2}+y ^{2}-2xy=1\]
let me denote dx/dy=x' since you want to derivate with respect to y
\[2xx' +2y-2(x'y+x)=0\]
so this is the first derivative with respect to y
did I answer your question
sorry for this late reply and ya u gave the answer :)
what is the derivative of 7955.6e^-0.0458/t
Would you mind posting each question separately? What is the derivative of e^t? of e^(-t)? To find the derivative of the function you've just given, you'll have to use (1) the rule for the derivative of e^u and (2) the chain rule, since your exponent is in itself a function, and (3) the power rule (since 1/x is a power of x.
(7955.6)e^(-0.0458/t)
Don't understand. Was that meant to be a question? an answer? a review? Please always include instructions or explanations with expressions such as this one. I'd be glad to help you, but do hope you'll answer my questions first.
this is a question and its like this v = (7955.6)e^(-0.0458/t) and i have to find dv/dt it means v' (v's derivation)
I asked you earlier: What is the derivative of e^t? of e^(-t)? To find the derivative of the function you've just given, you'll have to use (1) the rule for the derivative of e^u and (2) the chain rule, since your exponent is in itself a function, and (3) the power rule (since 1/x is a power of x. Mind answering? Then I'd be in a better position (as well as more willing) to address your question about differentiating v(t).
Example: If y = 5 e^(2t), the derivative would be dy/dt = 5 e^(2t) * 2 = 10*e^(2t). In obtaining this result, I used: (1) the rule for the derivative of e^u and (2) the chain rule, since your exponent is in itself a function, and (3) the power rule. Hope this makes sense to you; if not, ask questions. Better to learn from a simple example than from a more complicated problem.
Join our real-time social learning platform and learn together with your friends!