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Unformatted text preview: Math 9A Quiz 4 Solution
Name: 1. Let y= Find dy/dx. Solution: Use the chain rule: d dx x2 1 +x 8 x
4 1 x2 +x 8 x 4 . = 4 = 4 x2 1 +x 8 x 1 x2 +x 8 x 3 d dx x2 1 +x 8 x . 3 x 1 +1+ 2 4 x 2. For the parametric curve given by the equations x = 2 cos t, y = 2 sin t, find an equation for the tangent line to the curve at the point defined by t = /4. Solution: The point (x0 , y0 ) corresponding to the parameter value t = /4 is (2 cos(/4), 2 sin(/4 ( 2, 2). The tangent line will look like y  y0 = m(x  x0 ), where the slope m is given by the derivative of the function at the specificed point. To find this slope we use the derivative formula dy dy/dt = , dx dx/dt and evaluate it at the parameter value t = /4. We have dy 2 cos t dy 2( 2/2) = , and so  = = 1. dx 2 sin t dx t=/2 2( 2/2) Therefore the equation for the tangent line is y  2 = (x  2), or y = x + 2 2. 3. Find dy/dx and then d2 y/dx2 for the equation y 2 = x2 + 2x. Solution: To find dy/dx, use implicit differentiation... y2 d/dx = x2 + 2x
d/dx dy 2y dx = 2x + 2 ... and then solve for dy/dx: dy/dx = 2x + 2 . 2y Now, to find d2 y/dx2 , use implicit differentiation again. For the expression on the right, you will need the quotient rule! dy/dx
d/dx = 2x+2 2y d2 y/dx2 = d/dx(2x+2)2y+(2x+2)d/dx(2y) (2y)2 d/dx = Substituting in the expression dy/dx = d2 y/dx2 = 4y+(2x+2)(2dy/dx) 4y 2 2x+2 2y and simplifying gives us 4y + 4(x + 1)3 . 8y 3 4. Find Find dy/dx for the equation x = tan y. Solution: Again, use implicit differentiation: x
d/dx = tan y sec2 y dy/dx. d/dx 1 = Solving for dy/dx gives us dy/dx = 1/ sec2 y = cos2 y. ...
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This note was uploaded on 04/16/2008 for the course MATH 9A taught by Professor Apoorva during the Winter '07 term at UC Riverside.
 Winter '07
 APOORVA
 Chain Rule, The Chain Rule

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