Jaweria
  • Jaweria
Find dy/dx: 4x^2+3xy^2-6x^2y=y^3.
Mathematics
chestercat
  • chestercat
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abb0t
  • abb0t
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
Jaweria
  • Jaweria
ahan ok
abb0t
  • abb0t
Remember: \[3xy^2 = 3x \frac{ d }{ dx}y^2+y^2\frac{ d }{ dx }3x\]

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Jaweria
  • Jaweria
I m little bit confuse here
cwrw238
  • cwrw238
what you are doing is treating y as a function of x (which is implied in the expression)
abb0t
  • abb0t
Of course!
Jaweria
  • Jaweria
thanks :-)
abb0t
  • abb0t
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.
Jaweria
  • Jaweria
ok
abb0t
  • abb0t
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: \[\frac{ d }{ dx }y^3 = 3y^2\frac{ dy }{ dx } \]

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