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Basic Concepts of Differential and Integral Calculus

Introduction to Differentiation

Objective After going through this lesson, you shall be able to understand the following concepts. Differentiation First Principle of Derivative Standard Formulae of Derivative Algebra of Derivatives Introduction There are many physical phenomena that involve changing quantities and their rates of change. Let us take a very simple example. The rate of change of distance with respect to time gives us speed. Here, the distance is changing with respect to time. This rate of change is the speed. In the same manner, the number of bacteria in a bacteria culture, rate of flow of blood through the heart and many other phenomena involve the rate of change. The subject of differential calculus is concerned with the rate of change of a dependent variable with respect to an independent variable. Definition of Differentiation Suppose we have a function y = f (x) and we want to find the rate of change of the function with respect to variable x, assuming that y is the dependent variable and x is the independent variable. Here, comes the role of differentiation. We can find the rate of change of a function with respect to some variable with the help of differentiation. Differentiation is the process of finding the rate of change of one variable with respect to another. It is the limiting value of the ratio of change in the function to a small change in the independent variable. First Principle of Derivative Suppose y = f (x) is a real-valued function and h is a small increment in the value of x. Then, their would be a corresponding change in the value of y or f (x). This change in the value is f (x + h) − f (x). The derivative of f {denoted by or } is defined by This definition of derivative is called the first principle of derivative. Suppose f is a real-valued function and a is a point in the domain of definition. If the limit exists, then it is called the derivative of f at a. The derivative of f at a is denoted by.

 

The derivative of f (x) is also known as the differential coefficient of f (x) with respect to x.  

Example 1: Find the derivative of from the first principle of derivative. Solution Let

Thus, the derivative of is . Example 2: Using the first principle, find the derivative of .

Solution

Example 3: Using the first principle of derivative, find the derivative of (ax2 + bx + c)n with respect to x, where a, b, c and n are constants. Solution Let f (x) = (ax2 + bx + c)n

Some Standard Formulae of Derivative 1. ddxx=1 Proof: Here,  f(x) = x. Now, using the first principle of derivative, i.e. , we get ddxx=limh→0fx+h-fxh            =limh→0x+h-xh            =limh→0hh            =1 2. ddxc=0, where c is a constant The derivative of a constant is always 0. Proof: Here,  f(x) = c. Now, using the first principle of derivative, i.e. , we get ddxc=limh→0fx+h-fxh             =limh→0c-ch             =limh→00h             =0 Example: ddx5=0, ddxπ=0 3. ddxxn=nxn-1, where n is any constant Proof: Here,  f(x) = xn. Now, using the first principle of derivative, i.e. , we get ddxxn=limh→0fx+h-fxh              =limh→0x+hn-xnh      .....1 Put x + h = p⇒h = p − x If h → 0, then p → x. Putting the value of x + h and h in (1), we get ddxxn=limp→xpn-xnp-x              =nxn- 1 Example: ddxx5=5x5-1=5x4 ddx1x3=ddxx-3=-3x-3-1=-3x-4=-3x4 4. ddxex=ex Proof: Here,  f(x) = ex. Now, using the first principle of derivative, i.e. , we get ddxex=limh→0fx+h-fxh              =limh→0ex+h-exh               =limh→0ex·eh-exh               =limh→0exeh-1h               =exlimh→0eh-1h              =ex ×1              =ex 5. ddxlogx=1x Proof: Here, f(x) = logx. Now, using the first principle of derivative, i.e. , we get ddxlogx=limh→0fx+h-fxh                 =limh→0logx+h-logxh                  =limh→0logx+hxh                  =limh→0log1+hxh                 =limh→01hlog1+hx        .....1 Put hx=p ⇒ h = px If h → 0, then p → 0. Putting the value of x + h and h in (1), we get ddxlogx=limp→01pxlog1+p                  =1xlimp→0log1+pp                  =1x×1                  =1x 6. ddxax=ax logea Proof: Here,  f(x) = ax. Now, using the first principle of derivative, i.e. , we get ddxax=limh→0fx+h-fxh              =limh→0ax+h-axh               =limh→0ax·ah-axh               =limh→0axah-1h               =axlimh→0ah-1h              =ax logea Example: ddx5x=5xloge5 Algebra of Derivatives Let f(x) and g(x) be two differentiable functions. Suppose c is a constant, then the following cases hold true. 1. Differentiation of the product of a function and a constant        ddxc.fx=c.ddxfx     The derivative of the constant times a function is equal to the constant times the derivative of the function. 2. Differentiation of the sum of two functions      ddxfx+gx=ddxfx+ddxgx     The derivative of the sum of two functions is equal to the sum of their derivatives. 3. Differentiation of the difference of two functions     ddxfx-gx=ddxfx-ddxgx     The derivative of the difference of two functions is equal to the difference of their derivatives. 4. Differentiation of the product of two functions     ddxfx·gx=fxddxgx+gxddxfx     Derivative of the product of two functions = First function × Derivative of the second function + Second function × Derivative of the first function     This result is called the product rule of differentiation. 5. Differentiation of the division of two functions     ddxfxgx=gxddxfx-fxddxgxgx2     Derivative of the quotient of two functions = Function in denominator ×Derivative of function in numerator-Function in numerator ×Derivative of function in denominatorFunction in denominator2     This result is called the quotient rule of differentiation. Example 4: Find the derivative of . Solution

Example 5: Find the value(s) of x for which the product of (1 − x) and the derivative of is − 255. Solution

= −255

⇒ = −255

⇒ (1 − x) (1 + x + x2 + x3 + … + x7) = −255

⇒ 1 − x8 = −255      [an − xn= (a − x) (xn−1 + xn−2 a + xn−3 a2 + ... + x an−2 + an−1)]

⇒ x8 = 256 = (±2)8

⇒ x = ±2

Thus, the required values of x are 2 and −2.

Example 6: Find the derivative of 3x2x+logx. Solution The given function is 3x2x+logx. This function has been expressed as the product of the two functions 3x and 2x+logx. So, we can find the derivative of the given function using the product rule. ddx3x2x+logx=3xddx2x+logx+2x+logxddx3x                                 =3xddx2x+ddxlogx+2x+logxddx3x                                 =3x2ddxx+ddxlogx+2x+logxddx3x                                 =3x2×1+1x+2x+logx×3xloge3         &#

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