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abb0t Group TitleBest ResponseYou've already chosen the best response.1
Using implicit differentiation with respect to x. This means that you take the derivative of everything, even y, but everytime you take the deriviative of y, you multiply it by y'. For instance:\[\frac{ dy }{ dx }(x+y) = 1+\frac{ dy }{ dx } = 1+y'\] then, using your algebra skills, rearrange to find dy/dx
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
Remember: \[3xy^2 = 3x \frac{ d }{ dx}y^2+y^2\frac{ d }{ dx }3x\]
 one year ago

Jaweria Group TitleBest ResponseYou've already chosen the best response.0
I m little bit confuse here
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
what you are doing is treating y as a function of x (which is implied in the expression)
 one year ago

Jaweria Group TitleBest ResponseYou've already chosen the best response.0
thanks :)
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
Start by taking the derivative as you normally would for the left side of the function. And like @cwrw238 pointed out, you're treating y as a function of x.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
If it helps, you can break it down to see it more clearly: \[\frac{ d }{ dx }4x^2 =\] \[\frac{ d }{ dx }3xy^2 = (3x \times \frac{ d }{ dx }y^2)+(y^2\frac{ d }{ dx }3x) = [3x \times 2y \frac{ dy }{ dx }]+[y^2 \times 3]\] \[\frac{ d }{ dx }6x^2y = \] NOW, the right side: \[\frac{ d }{ dx }y^3 = 3y^2\frac{ dy }{ dx } \]
 one year ago
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