4 problems I can't figure out: 1) (x^2)(y^2) - y = x 2) 2sinx(cosy) = 1 3) sinx = x(1+tany) 4) y = sin(xy) Please don't simplify or try to isolate y from the beginning. =) Only simplify to isolate dy/dx
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Originally Posted by helpplz 4 problems I can't figure out: 1) (x^2)(y^2) - y = x product rule with the $\displaystyle \textcolor{red}{x^2y^2}$ term 2) 2sinx(cosy) = 1 product rule again ... 3) sinx = x(1+tany) product rule on the right ... 4) y = sin(xy) product rule with the xy argument what seems to be your problem with using the product rule for derivatives?
I'm not sure why I just don't know where to start..I think I just don't know how to apply product rule with the y's..
Originally Posted by helpplz I'm not sure why I just don't know where to start..I think I just don't know how to apply product rule with the y's.. (1st function)(derivative of the 2nd) + (2nd function)(derivative of the 1st) like so ... $\displaystyle \frac{d}{dx}(x^2y^2) = x^2 \cdot 2y \frac{dy}{dx} + y^2 \cdot 2x$
Hm, but why is it 2y(dy/dx) instead of just 2(dy/dx)?
Originally Posted by helpplz Hm, but why is it 2y(dy/dx) instead of just 2(dy/dx)? because it was $\displaystyle y^2$ to begin with ... not $\displaystyle 2y$. you still have to use the power rule.
Originally Posted by skeeter because it was $\displaystyle y^2$ to begin with ... not $\displaystyle 2y$. you still have to use the power rule. So if it was something like y^3, the derivative would be 3y(dy/dx)^2 ?
Originally Posted by helpplz So if it was something like y^3, the derivative would be 3y(dy/dx)^2 ? no $\displaystyle \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$
Originally Posted by skeeter no $\displaystyle \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$ Ah, okay, gotcha. I'm gonna do a random problem just to make sure I'm doing it right: x^2y^2 -2x = 3 x^2(2y(dy/dx)) + y^2(2x) = 5 dy/dx = 5-y^2(2x)/ x^2(2y) That right?
Originally Posted by helpplz Ah, okay, gotcha. I'm gonna do a random problem just to make sure I'm doing it right: x^2y^2 -2x = 3 x^2(2y(dy/dx)) + y^2(2x) = 5 not quite ... $\displaystyle \textcolor{red}{x^2 \cdot 2y \frac{dy}{dx} + y^2 \cdot 2x - 2 = 0}$ the derivative of a constant is 0.
Originally Posted by skeeter the derivative of a constant is 0. Oh geez, I must have forgotten that on at least 15 problems...Thanks sooo much for correcting me. =) And for all your help!
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