By R. M. Johnson

This lucid and balanced creation for first yr engineers and utilized mathematicians conveys the transparent knowing of the basics and functions of calculus, as a prelude to learning extra complex features. brief and primary diagnostic routines at bankruptcy ends try out comprehension sooner than relocating to new fabric.

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**Extra info for Calculus: Introductory Theory and Applications in Physical and Life Science**

**Sample text**

Evaluate the following. (i) / sin 20 liml — θ-ο\ Θ (ii) lim r-«-o\ tan t/ (iii) lim χ-*π/2\(π/2)— x t cos* Sketch the graph of the function y = fix) where Ί+jc, Λ:<0, fix) •2+x, x>0. Evaluate lim {/(*)} and lim {/(*)}. ) 6. The function /(*) is said to be continuous at x = a if lim {/(*)} exists and is equal to /(a). ) (i) Confirm that the function defined in problem 5 is continuous for all x except x = 0. (ii) Sketch roughly the graph of y = \/x2 and explain why it is not continuous when* = 0.

1. Sec. 1 d dv du — (uv) = u — + V-. 1a) v(duldx) — u(dv/dx) 3 dx\v/ v · ( 2 - 2a > Proof Let Ax be a small change in x resulting in changes Au in u, Av in v and Ay in y where y = uv. Therefore, y + Ay = (M + ΔΜ) (V + Av) = uv + u Av + v Au + Au Av. Thus, Ay = u Av + v Au + Au Av and Ay Av — = u l· Δχ Δχ v Au Δχ 1- Av ΔΜ — . Δχ Now let Δχ -»■ 0 (also Δ« -* 0, since u is differentiable), and by the definition of a derivative we obtain dy dv du — =u \- v— . 1a) to the product u =yv to give du dv dx dx dy 1 I du u dv dx v \dx v dx — =y l· dy v— .

As 13. 7 sin 30/. Determine the maximum value of the velocity of the component. 14. Given that y = 3x2 and àx/dt = 4 evaluate dy/dt when x = 1. Answers 1. (0 ί · (Ü) 2. 3. 4. 2. 32. 3. T· Sec. 5] 6. y + 4x-ll=0. 7. 16. 8. 9 + 3 Γ 3 / 2 - 2 Γ 3 . 9. - 1 . ' 10. - 3 . 5 , 11. 4π. 12. —2 cos s sins. 13. 21 m/s. 14. 24. 1 INTRODUCTION In Chapter 1 we saw that the derivative of the sum (or difference) of two functions was simply the sum (or difference) of their derivatives. However, the derivative of the product (or quotient) of two functions is not the product (or quotient) of the derivatives.