√ g) y = x/ 1 − x2 , dy/dx = (1 − x2 )−3/2 , 1 + y 2 = 1/(1 − x2 ). Thus d 1 dy/dx = (1 − x2 )−3/2 (1 − x2 ) = √ tan−1 y = dx 1 + y2 1 − x2 Why is this the same as the derivative of sin−1 x? √ √ h) y = 1 − x, dy/dx = −1/2 1 − x, 1 − y 2 = x. Thus, d dy/dx = sin−1 y = dx 2 1 − y2 x(1 − x) −1
5A-4 a) y ′ = sinh x. A tangent line through the origin has the equation y = mx. If it meets the graph at x = a, then ma = cosh(a) and m = sinh(a). Therefore, a sinh(a) = cosh(a) . b) Take the difference: F (a) = a sinh(a) − cosh(a) Newton’s method for finding F (a) = 0, is the iteration an+1 = an − F (an )/F ′ (an ) = an − tanh(an ) + 1/an With a1 = 1, a2 = 1.2384, a3 = 1.2009, a4 = 1.19968. A serviceable approximation is a ≈ 1.2 (The slope is m = sinh(a) ≈ 1.5.) The functions F and y are even. By symmetry, there...
No comments