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Find the derivative of : Y = (2x-5)(4 – x) -1

Find the derivative of : Y = (2x-5)(4 – x) -1. Find the derivative of: S = csc 5 (1- t + 3t ²). Find the derivative of: Y = -1 15(15t – 1) ³. Find the derivative of: Y = 3 (5x ² + sin2x ) 3/2.

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Find the derivative of : Y = (2x-5)(4 – x) -1

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  1. Find the derivative of : Y = (2x-5)(4 – x)-1

  2. Find the derivative of: S = csc5(1- t + 3t²)

  3. Find the derivative of: Y = -1 15(15t – 1)³

  4. Find the derivative of: Y = 3 (5x² + sin2x )3/2

  5. Find an equation for the line in the xy plane that is tangent to the curve at the point corresponding to the given value of t. Also, find the vlaue of d²y/dx² X = 1 + 1/t² y= 1 – 3/t t = 2

  6. Using implicit differentiation find dy/dx of 5x4/5 + 10y6/5 = 15

  7. Using implicit differentiation find the equations for the lines that are tangent and normal to the curve at the given point. (y – x)² = 2x + 4 (6,2)

  8. Find • Δf = f(x0 + dx) – f(x) • Value of estimate df = f′(x0)dx • Approximate error │Δf - df| • For x³-2x + 3 x0 = 2 dx = .1

  9. Find dy of 2y2/3 + xy – x = 0

  10. Find the linearization of f(x) = √1 + x + sinx - .5 at x = 0

  11. At time t seconds, the positions of 2 particles on a coordinate line are s1= 3t³ - 12t² + 18t + 5 meters, and s2 = -t³ + 9t² - 12t meters. When do the particles have the same velocity?

  12. The volume of a cube is increasing at a rate of 1200 cm³/min at the instant its edges are 20 cm long. At what rate are the lengths of the edges changing at that instant.

  13. On earth, you can easily shoot a paperclip 64 feet straight up in the air with a rubber band. In t seconds after firing the paperclip is • S = 64t – 16t² feet above your hand. • How long does it take the paperclip to reach its maximum height? With what velocity does it leave your hand? • On the moon, the same acceleration will send the paperclip to a height of s = 64t – 2.6t² feet in t seconds. About how long will it take the paperclip to reach its maximum height and how high will it go?

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