 Hello and welcome to the session. Let us understand the following question which says a balloon which always remains spherical has a variable diameter 3 by 2 multiplied by 2x plus 1. Find the rate of change of its volume with respect to x. Now let us proceed on to the solution. The fifth I meter of the balloon is 3 by 2 2x plus 1. Hence r is equal to 1 by 2 multiplied by 3 by 2 multiplied by 2x plus 1. Therefore r is equal to 3 by 4 multiplied by 2x plus 1. Let us name it as 1. We know volume of the balloon that is sphere is given by v is equal to 4 by 3 pi r cube. Now substituting the value of r from 1 in v, we get v is equal to 4 by 3 pi multiplied by 3 by 4 multiplied by 2x plus 1 the whole cube which is equal to 4 by 3 pi multiplied by 3 cube is equal to 27 divided by 4 cube is equal to 64 multiplied by 2x plus 1 the whole cube which is equal to 4 by 3 multiplied by 27 by 64 pi multiplied by 2x plus 1 the whole cube which is equal to 27 gets cancelled by 3 and we get here 9 and 4 gets cancelled by 64 and we get here 16. So it is equal to 9 pi by 16 2x plus 1 whole cube. Now differentiating it with respect to x we get dv by dx is equal to 9 pi by 16 multiplied by 3 multiplied by 2x plus 1 the whole square multiplied by d by dx of 2x plus 1 which is equal to 9 pi by 16 multiplied by 3 multiplied by 2x plus 1 whole square multiplied by 2. Now 16 gets cancelled by 2 and we get here 8. So it is equal to 9 is a 27 pi by 8 multiplied by 2x plus 1 the whole square that is dv by dx. Hence volume is changing at the rate of 27 pi by 8 multiplied by 2x plus 1 the whole square. This is the required answer. I hope you understood this question by a nice day.