First considerx+1x2-xxx+x+xLet x=tx=t2x2=t4t+1t4-tt3+t2+t=tt+1t3-1tt2+t+1=t+1t-1t2+t+1t2+t+1=t+1t-1=t2-1=x-1Considery=x+1x2-xxx+x+x+115cos2x-5cos3xy=x-1+115cos2x-5cos3xdydx=d dxx-1+115cos3xd dxcos2x-5+115cos2x-5d dxcos3xdydx=1+115cos3x2cos xd cos xdx+115cos2x-53cos2x d cos xdxdydx=1+115cos3x-2cos x sin x +115cos2x-53cos2x -sin xdydx=1-215cos4x sin x-315cos4x sin x+1515cos2x sin xdydx=1-515cos4x sin x+cos2x sin xdydx=1-515cos2x sin x cos2x+cos2x sin xdydx=1-515cos2x sin x 1-sin2x+cos2x sin xdydx=1-515cos2x sin x +515 cos2x sin3x+cos2x sin xdydx=1+515 cos2x sin3x+1015cos2x sin xdydx=1+515 cos2x sin3x+1015cos2x sin xdydx=1+13 cos2x sin3x+23cos2x sin xPlease check you question again

Question 29

Question 29 29. Given v = and the points on the cur'. e to' ANSWERS

F i r s t c o n s i d e r \frac{(\sqrt{x} + 1) (x^{2} - \sqrt{x})}{x \sqrt{x} + x + \sqrt{x}} L e t \sqrt{x} = t x = t^{2} x^{2} = t^{4} \frac{(t + 1) (t^{4} - t)}{t^{3} + t^{2} + t} = \frac{t (t + 1) (t^{3} - 1)}{t (t^{2} + t + 1)} = \frac{(t + 1) (t - 1) (t^{2} + t + 1)}{(t^{2} + t + 1)} = (t + 1) (t - 1) = t^{2} - 1 = x - 1 C o n s i d e r y = \frac{(\sqrt{x} + 1) (x^{2} - \sqrt{x})}{x \sqrt{x} + x + \sqrt{x}} + \frac{1}{15} (\cos^{2} x - 5) \cos^{3} x y = x - 1 + \frac{1}{15} (\cos^{2} x - 5) \cos^{3} x \frac{d y}{d x} = \frac{d}{d x} (x - 1) + \frac{1}{15} \cos^{3} x \frac{d}{d x} (\cos^{2} x - 5) + \frac{1}{15} (\cos^{2} x - 5) \frac{d}{d x} \cos^{3} x \frac{d y}{d x} = 1 + \frac{1}{15} \cos^{3} x (2 \cos x \frac{d \cos x}{d x}) + \frac{1}{15} (\cos^{2} x - 5) 3 \cos^{2} x \frac{d \cos x}{d x} \frac{d y}{d x} = 1 + \frac{1}{15} \cos^{3} x (- 2 \cos x \sin x) + \frac{1}{15} (\cos^{2} x - 5) 3 \cos^{2} x (- \sin x) \frac{d y}{d x} = 1 - \frac{2}{15} \cos^{4} x \sin x - \frac{3}{15} \cos^{4} x \sin x + \frac{15}{15} \cos^{2} x \sin x \frac{d y}{d x} = 1 - \frac{5}{15} \cos^{4} x \sin x + \cos^{2} x \sin x \frac{d y}{d x} = 1 - \frac{5}{15} \cos^{2} x \sin x \cos^{2} x + \cos^{2} x \sin x \frac{d y}{d x} = 1 - \frac{5}{15} \cos^{2} x \sin x (1 - \sin^{2} x) + \cos^{2} x \sin x \frac{d y}{d x} = 1 - \frac{5}{15} \cos^{2} x \sin x + \frac{5}{15} \cos^{2} x \sin^{3} x + \cos^{2} x \sin x \frac{d y}{d x} = 1 + \frac{5}{15} \cos^{2} x \sin^{3} x + \frac{10}{15} \cos^{2} x \sin x \frac{d y}{d x} = 1 + \frac{5}{15} \cos^{2} x \sin^{3} x + \frac{10}{15} \cos^{2} x \sin x \frac{d y}{d x} = 1 + \frac{1}{3} \cos^{2} x \sin^{3} x + \frac{2}{3} \cos^{2} x \sin x P l e a s e c h e c k y o u q u e s t i o n a g a i n

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Question 29 Question 29 29. Given v = and the points on the cur'. e to' ANSWERS

Question 29

Question 29 29. Given v = and the points on the cur'. e to' ANSWERS