Show that y = log 1 + x - 2 x 2 + x , x > - 1 , is an increasing function of x throughout its domain.

For the given function y = log 1 + x - 2 x 2 + x On differentiating both sides w.r.t. x we get d y d x = d d x log ( 1 + x ) - 2 x 2 + x d y d x = d d x log 1 + x - d d x 2 x 2 + x d y d x = 1 1 + x - 2 + x d d x 2 x - 2 x d d x 2 + x 2 + x 2 ∵ d d x a b = b d d x a - a d d x b b 2 d y d x = 1 1 + x - 2 + x 2 - 2 x 2 + x 2 d y d x = 1 1 + x - 4 + 2 x - 2 x 2 + x 2 d y d x = 1 1 + x - 2 2 + x 2 d y d x = 2 + x 2 - 4 1 + x 1 + x 2 + x 2 d y d x = x 2 + 4 x + 4 - 4 - 4 x 1 + x 2 + x 2 d y d x = x 2 1 + x 2 + x 2 d y d x = x 2 2 + x 2 1 1 + x For y to be increasing d y d x > 0 , Then x 2 2 + x 2 1 1 + x > 0 If 1 + x > 0 ⇒ x > - 1 ∵ x 2 + x 2 is always + v e . Then, f ' x > 0 Hence, The function is increasing function of x throughout its domain.

The surface area of a spherical balloon is increasing at the rate of 2 cm 2 / sec . At what rate the volume of the balloon is increasing when the radius of the balloon is 6 cm .

Let S be the surface area of a spherical balloon and r is the radius of the balloon. Given, d S d t = 2 cm 2 / sec and r = 6 cm . We know that S = 4 πr 2 Differentiating both sides with respect to time t , we get: d S d t = 8 πr · d r d t ⇒ d S d t = 8 π × 6 × d r d t ⇒ 2 = 48 π · d r d t ⇒ d r d t = 1 24 π cm / sec We know that V = 4 3 πr 3 Differentiating both sides with respect to time t , we get: d V d t = 4 3 π × 3 r 2 × d r d t = 4 π × 36 × 1 24 π ∵ r = 6 cm = 6 cm 3 / sec Hence, the volume of the balloon is increasing at the rate of 6 cm 3 / sec .