There is a solution to 97 Exercise Problems in 11th math for the Tamilnadu syllabus.In 11th standard this is an important chapter.If a student wants to get good marks then they should master this chapter.Special focus is given to the tools that are developed based on the derivatives that are applied in real life and other related concepts in the chapter.The average of a rate is x and if the instance happens over a period of time.

They will keep the averate rate as x.For example if a student wants to score 90 percent on all the subjects.He/she needs to score higher than 90 in some subjects as he/she may not score as high in other subjects.The time rate of change of score is defined by the number of subjects and the average rate of scoreAny moving object the same is applicable.

A person can run at a speed of 20 km/h.The distance traveled divided by the time taken shows the rate of speed.If the runner is 3 km from the start the speed is 3/6*60.This is how much it would take to travel 30 km/HR.This is not a true measure for rate.

The rate of speed will go up toThis is equivalent to 60 km/hr.There are four major problems in math.In the section that will follow we will see first two details.When a circle is formed the tangent to the circle will cross the border of the circle.

There are scenarios where the curve only passes one time through the border.There are other occurances in the curve where the tangent passes through multiple points.The easiest way to find the slope of the line that passes through two points in a curve is to calculate the tangent of the curve.You can find the slope of the curve with differential quotient.It is divided into two parts -Delta y andDelta x.

The slope of the curve is also known as the slope of the angle line.A position function is used for calculating the velocity.The change in distance would be divided by the change in time to make it simpler.It would be simpler to calculate thevelocity using the position function if we measure the time and distance at two points in time.The logic of differentiating is that y is a function of x

With regards to x we will differentiate y.This result will be called dy/dx.We will get f'(x) if we differentiate f(x)(2)(x)(2)(x)(2)(x)(2)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(x)(2)(xSimilarly y' can be written as dy/x.We're going to see examples of differentiating y and x.

10 x9 is the result of x10.There are different willluts in 20 x19.-2 x-4 is the result of x-1 differentiating.In -11x-12 the differentiating x-11 will be different.Differentiating x1/2 will result in 1/2x-1/2.

When we differentiate y with respect to x we will get dy/dx which is 10 x9 + 7 x6 + 5 x4 and 3 x2.They will get zero if we differentiate a constant.It is called as constant if there is no x in the element.6x0 when we differentiate we get 6*0*x-1 which will be zero.The difference between 5 + x3 will be zero.