#$&* course Mth 271 11/12 830p *********************************************
.............................................
Given Solution: `a The derivative of x^2 with respect to x is 2 x. The derivative of y^3 with respect to x is 3 y^2 dy/dx. You can see this by realizing that since y is implicitly a function of x, y^3 is a composite function: inner function is y(x), outer function f(z) = z^3. So the derivative is y'(x) * 3 * f(y(x)) = dy/dx * 3 * y^3. So the derivative of the equation is 2 x - 3 y^3 dy/dx = 0, giving 3 y^2 dy/dx = 2 x so dy/dx = 2 x / ( 3 y^2). At (2,1), we have x = 2 and y = 1 so dy/dx = 2 * 2 / (3 * 1^2) = 4/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):It seems so much simpler to take the derivative of the x term divided by the derivative of the y term, since that’s what you have before you substitute.
.............................................
Given Solution: `a The derivative of the equation is 2 x - 3 y^2 dy/dx = 0. Solving for dy/dx we get dy/dx = 2x / (3 y^2). At (-1,1) we have x = 1 and y = 1 so at this point dy/dx = 2 * -1 / (3 * 1^2) = -2/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): That was the last question that I left out on the previous problem. I did not see where the negative was used in arriving at the answer. It just seemed to disappear, but it didn’t here. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q 6 7th edition 2.7.42 (was 2.7.36) p=`sqrt( (500-x)/(2x)) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [(500-x)/(2x)]= ^½ (½)[(500-x)/(2x)]^(-3/2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a You could apply implicit differentiation to the present form, and that would work but it would be fairly messy. You have lots of choices for valid ways to rewrite the equation but I would recommend squaring both sides and getting rid of denominators. You get p^2 = (500-x) / (2x) so 2x p^2 = 500-x and 2x p^2 + x - 500 = 0. You want dx/dp so take the derivative with respect to p: 2x * 2p + 2 dx/dp * p^2 + dx / dp = 0 (2 p^2 + 1) dx/dp = - 4 x p dx / dp = -4 x p / (2p^2 + 1) ** STUDENT QUESTION: could you explain that differiant I not sure what happens to the middle x INSTRUCTOR RESPONSE Remember how implicit differentiation works, with on variable being considered as a function of the other. In this case x is considered to be a function of p. One way to see it: if ' represents derivative with respect to p, then (2 x p^2) ' = 2 ( x p^2) ' = 2 ( x ' p^2 + x * (p^2)' ) = 2 ( x ' p^2 + x * (2 p) ) = 2 x ' p^2 + 4 x p. So the equation 2x p^2 + x - 500 = 0 gives us (2 x p^2) ' + x ' + 0 = 0, or 2 x ' p^2 + 4 x p + x ' = 0. If we use d/dp notation, this is 2 dx/dp + p^2 + dx/dp = 0. Remember that the derivative is with respect to p, not x, so x ' is x ' = dx/dp, not x ' = 1 (as it would be if we were taking a derivative with respect to x). Another way to notate it: The equation is 2 x p^2 = 500 - x. We take the derivative of both sides with respect to p. So the derivative of x is dx/dp. The derivative of p^2 with respect to p is just 2 p. The derivative of x p^2 is, by the product rule, derivative of x p^2 = (derivative of x) * p^2 + x * (derivative of p^2), which works out to dx/dp * p^2 + x * (2 p). Of course, that means that the derivative of 2 x p^2 is 2 ( dx/dp * p^2 + x * (2 p) ) = 2 dx/dp * p^2 + 4 x p. So the derivative with respect to p of 2 x p^2 + x - 500 is (2 dx/dp * p^2 + 4 x p) + dx/dp + 0 or just (2 dx/dp * p^2 + 4 x p) + dx/dp, and our equation is 2 dx/dp * p^2 + 4 x p + dx/dp = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I really don’t understand just substituting dx/dp for the derivative instead of just finding the derivative. I read through the entire explanation and I don’t really understand it.