Samacheer Kalvi 11th Maths Solutions எல்லாம் ஒரே இடத்தில் இங்கு உங்களுக்காக தருகின்றோம். ஒரு புதுமையான முறையில் இங்கு நீங்கள் படங்களை போர்டு எழுத்துக்கள் மூலம் கற்கலாம்.

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Let us consider a runner running at a speed of 20km/hr. At any point in time the measure of rate of speed is distance travelled divided by the time taken. For example, at 6 minutes if the runner is at 3 km from starting of the run, the speed would be 3/6*60. This is equal to 30km/hr. However this is not a true measure of rate.

If the runner is at 5km at 8minutes then the current rate of speed would be derived by measuring from the last known speed and time. The current rate of speed will be (5-3)/(8-6)*60. This is equal to 60km/hr.

In calculus mathematicians solve the following four major problems. We will be seeing first two in details in the coming section.

For a circle the tangent to a circle will cross the border of the circle which will be perpendicular to the radius that goes through that point. But the tangent of curve is not so easy. There are scenarios where in a tangent of a curve passes only once through the border of the curve. But there are other occurances where in the tangent might pass through multiple points in the curve. The easy method to calculate the tangent of a curve at any point is to find the slope of the line that passes through two points in that curve.
Differential quotient is used for finding the slope of the curve. It is delta y divided by delta x. i.e change in y concerning the change in x. The slope of the tangent line is also known as the slope of the curve.

The velocity is calculated using position function. This would be simplified by having ration of the change in distance divided by change in time. we need to measure the time and distance at two point in time and then it would be simpler to calculate the velocity using the position function.

The logic of differentiation is y is always a function of x. so we can write this as y = f(x). Now we will differentiate y with respect to x. This will result in dy/dx. If we differentiate f(x), then we will get f'(x). Similarly dy/dx can be written as y'.

Let us see few examples of differentiating y with respect to x. If y is x^n, then differentiating it will give n * x^(n-1). x^10 differentiating will result in 10 x^9. x^20 differentiating will reslu\t in 20 x^19. x^-3 differentiating will result in -3 x^-4. Differentiating x^-11 will reslut in -11x^-12. Differentiating x^1/2 will result in 1/2x^-1/2. When differentiating x^-3/2 will result in -3/2 x -5/2.

if y = x^10 + x^7 + x^5 + x^3 then when we differentiate y with respect to x we will get dy/dx = 10 x^9 + 7 x^6 + 5 x^4 + 3 x^2

If we differentiate a constant, we will get zero. Any element without x is called as constant. 6x^0 when we differentiate we weill get 6*0*x^-1 which will result in zero.

Differentiating 5 + x^3 will result in 0 + 3 x^2. we need to differentiate only the variable x not the constant element.

Differentiating 5 x^3 will result in 5 * 3 x^2 = 15 x^2.

y=2 x^4 - 3 x^3 + 6 x^2 + 7 x + p - 5

Here if we differentiate y we will get, 6 x^3 - 9 x^2 + 12 x^1 + 7