# Calculus

Calculus is the study of derivatives. It is based on the rate of change for example if we consider velocity = $\frac{dx}{dt}$. Where, x is the distance, which keeps changing with time.

In the above example we see that velocity is different at different time intervals. Hence, we infer this as rate of change of time with respect to distance.

So we always try to say that "A variable is being differentiated with respect to some other variable"

## Calculus Formulas

We shall try to put some formulas that is being used regularly.

1. $\frac{d}{dx}$ ( xn) = nxn-1

2. $\frac{d}{dx}$ (sinx) =cosx

3. $\frac{d}{dx}$(cosx)=-sinx

4. $\frac{d}{dx}$(tanx) =sec2 x

5. $\frac{d}{dx}$(cotx) =-cosec2x

6. $\frac{d}{dx}$(secx) = secx tanx

7. $\frac{d}{dx}$(cosecx)= -cosecx cotx

8.
9.
10.

## Calculus Problems

Find the derivative of x2 + 3 sinx - tanx.

Answer: $\frac{d}{dx}$ ( x2) + 3 $\frac{d}{dx}$( sinx) - $\frac{d}{dx}$(tanx).

We know that the derivative of xn = n xn-1

so 2x2-1 = 2x

$\frac{d}{dx}$(sinx) = cosx

$\frac{d}{dx}$( tanx) = sec2 x.

Answer: 2x + 3cosx -sec2 x

Find the derivative of e3x - e 2x + cotx

Answer: $\frac{d}{dx}$ (e3x) -$\frac{d}{dx}$(e2x) + $\frac{d}{dx}$ (cotx)

: 3 e3x - 2 e 2x - cosec2 x

The distance travelled by the car is given by the equation X = 3t3-t2 find the velocity of the car at the instant t= 1sec

We all know that velocity = $\frac{dx}{dt}$

= $\frac{d}{dt}$ (3t3-t2)

= 3(3t2) - [2t].

= 9 t2 -2t

= 9[1] 2 -2[1]

= 9-2

= 7m/sec.

The distance travelled by the car is given by the equation X = 3t4-t2 find the velocity of the car at the instant t= 1sec

We all know that velocity = $\frac{dx}{dt}$

= $\frac{d}{dt}$ (3t4-t2)

= 4(3t3) - [2t].

= 12 t2 -2t

= 12[1] 2 -2[1]

= 12-2

= 10m/sec.

## Calculus Problem Solver

The distance travelled by the car is given by the equation X = t3-t4 find the velocity of the car at the instant t= 2sec

We all know that velocity = $\frac{dx}{dt}$

= $\frac{d}{dt}$ (t3t4)

= 3(t2) - (4t3).

= 3 t2 -4t3

= 3[2] 2 -4[2] 3

= 12-24

= -12 m/sec. [the vehicle is coming to halt]

The distance travelled by the car is given by the equation X = 3t4-t2 find the velocity of the car at the instant t= 2sec

We all know that velocity = $\frac{dx}{dt}$

= $\frac{d}{dt}$ (3t4-t2)

= 4(3t3) - [2t].

= 12 t2 -2t

= 12[2] 2 -2[2]

= 48-4

= 44 m/sec.

Next Chapters

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