📖 generic · CBSE Class 12th English Medium · MATHEMATCS PART-1 · Page 23question

A Note The reader may note that in Example 35, we have used first derivative · Part 12

Chapter 6: APPLICATION OF DERIVATIVES · MATHEMATCS PART-1

∞ ), i.e., when x > . In this case, we have x – > , x + > and x – > . So f ′ ( x ) > when x > . Thus, f is increasing in the interval ( , ∞ ).

Example Show that the function f given by f ( x ) = tan – (sin x + cos x ), x > is always an increasing function in , π     . Solution We have f ( x ) = tan – (sin x + cos x ), x > Therefore f ′ ( x ) = (cos sin ) (sin cos ) Fig . = cos sin sin (on simplification) Note that + sin x > for all x in , Therefore f ′ ( x ) > if cos x – sin x > f ′ ( x ) > if cos x > sin x or cot x > Now cot x > if tan x < , i.e., if Thus f ′ ( x ) > in , π     Hence f is increasing function in , Example A circular disc of radius cm is being heated. Due to expansion, its radius increases at the rate of .

cm/s. Find the rate at which its area is increasing when radius is . cm. Solution Let r be the radius of the given disc and A be its area.

Then A = π r A dt = dr r dt (by Chain Rule) Now approximate rate of increase of radius = dr = . dr t dt ∆= cm/s. Therefore, the approximate rate of increase in area is given by d A = A ( t dt ∆ = dr t ∆ = π ( . ) ( .

) = . π cm /s ( r = . cm) Example An open topped box

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