Introduction to Derivative of the Function

In the Class 12 chapter ‘Continuity and Differentiability’, we were introduced to the topic of the derivative of the function. This article will discuss the different derivatives of trigonometric functions, second-order derivative function, and derivative of functions in parametric form, along with other concepts. 

What is differentiable? 

• If the derivative of the function exists at all the points in a domain, then that function is said to be differentiable. 

Rules established as per algebra of derivatives 

• (x z)’ = x’ z’ 

• (x z)’ = x’ z + x z’ 

• xz’ = x’z-x z’z2 

Derivative of trigonometric functions 

The theorem of continuity- The theorem of continuity states that if a function is continuous at a point, it is also continuous. This theorem also states that every differentiable function is continuous. 

Table of derivatives of trig functions 

Example 1- Find the derivative of y = cos 2x – 2 sin x. 

dydx = (cos 2x)’ – (2 sin x)’

dydx = (-sin 2x) * (2)’ – 2(sin x)’ = -2 sin 2x – 2 cos x

dydx = -4 sin x cos x – 2 cos x 

dydx = -2 cos x (sin x + 1)  

Example 2 – Find the derivative of y = cos x – 13 cos3 x. 

dydx = (cos x – 13 cos3 x)’ 

dydx = (cos x)’ – ( 13 cos3 x)’ 

dydx = – sin x – ( 13 ×3 cos2 x) (cos x)’ 

dydx = -sin x – cos2 x (-sin x) 

dydx = – sin x + cos2 x sin x 

dydx = – sin x (1 – cos2 x) 

dydx = – sin x sin2 x 

dydx = – sin3 x 

Derivatives of inverse trigonometric functions

Table of derivatives of inverse trigonometric functions 

Example 1 – Find the derivative of y = sin-1 (x – 1). 

dydx = sin-1 (x – 1)’ 

∵ (sin-1 x)’ =  11-x2 

∴ dydx = 11-(x-1)2

dydx = 11-x2-1+2x

dydx = 1-x2+2x

dydx = 12x – x2

Example 2- Find the derivative of y = tan-1 x+1x-1

dydx = tan-1 x+1x-1′

∵ (tan-1 x)’ = 11+ x2 

dydx = 11+ x+1x-12x+1x-1′

dydx = x-1-(x+1)(x-1)2+ (x+1)2

dydx = x-1-x-1×2+1-2x+ x2+1+ 2x

dydx = -22(x2+1)

dydx = -1(x2+1) 

Example 3 – Find the derivative of y = cot-1 1×2 

dydx = (cot-1 1×2)’ 

Since, (cot-1 x) = – 11+ x2

Therefore, dydx = – 11+ 1×22 1×2′

dydx = – 11+ 1×4 (-2x-3) 

dydx = 2x4x4+1×3 

dydx = 2×1+ x4 

Derivative of composite functions 

• If you can write a function as f (g (x)), then that function is a composite. Therefore, a composite function refers to a function within another function. 
• For example, cos (x2) is a composite function because it can be expressed as f (x) = cos x and as g (x) = x2. Therefore, cos (x2) can be expressed as f (g (x)). In this function g (x) is the inner function whereas function f (x) is the outer function. 
• Let us also examine an example of a function that is not an example of a composite function. Sec (x) x2 is not a composite function. However, it is a product of two functions. 

Theorem of chain rule- The theorem of chain rules helps us determine the derivative of a composite function. The chain rule states that- dfdx= d(wou)dt ×dtdx=dwdsdsdt dtdx

Example 1- Using the chain rule, find the derivative of the following function- sin (x2) 

Since, the function is a composite of two functions, therefore, t = u(x) = x2 and v(t) = sin t. 

Then, dfdx= d(wou)dt ×dtdx=dwdsdsdt ×dtdx 

dfdx= dvdt ×dtdx = cos t 2x

Therefore, dfdx= 2x cos x2

Alternatively, we could have also solved the function in the following way- 

Y = sin (x2)

dydx = ddx sin (x2)

dydx = cos x2 ddx x2

dydx = 2x cos x2

Derivatives of implicit functions

• It is not necessary that functions are always expressed as y = f(x). 
• For example, let us consider the relationship between x and y in the following equations- 

1. x – 2y =  
2. x + cos 2xy – 3y = 0

While in equation 1, we can solve for y by rewriting the equation as y = x-   2. 

However, in the second equation, it is not very easy to solve for y. 

Therefore, in equations where the relationship between x and y is expressed in such a way that it is easy to solve for y, then the given function is an explicit function. So, equation 1 is an explicit function, whereas the equation between x and y has resulted in an implicit function. 

• Example 1- Find the derivative of y + sin y = cos x 

dydx + cos y. (ddx) = – sin x. (ddx)

After applying the chain rule to the above equation, we get 

dydx + cos y. (dydx) = – sin x

dydx (1 + cos y) = -sin x 

dydx = -sin x (1 + cos y)

Derivative of the function in parametric forms 

• There are times when the relationship between two variables is neither explicit nor implicit, and the relationship between them can only be expressed via a third variable. The third variable, in this case, is called a parameter. 
• With the help of parametric form, the relationship between two variables, x and y, can be established using two equations- 

Let’s assume that variable t is a parameter, therefore, x = x (t) and y = y (t). 

• By applying the chain rule to the derivative of the function in parametric forms, we get 

by dx= g'(t)f'(t) ∵ dydt=g’t and dxdt=f'(t) provided f’t≠0

Second-order derivative function 

• The second-order derivative function refers to the derivative of the first derivative of a given function. 

• The second order derivative of a function f(x) is denoted by f’’(x), D2y, y’’, or y2. 

• The second-order derivative is useful because it gives the user an idea of the shape of the given function on a graph. 

• ∵ dy dx = f ‘(x)

Therefore, ddx dy dx = d2ydx2 

• For example, Find the second order derivative of y = x3+ tan x

dy dx = 3×2 + sec2 x 

d2ydx2 = ddx (3×2 + sec2 x)

d2ydx2= 6x + 2 sec x. sec x. tan x = 6x + 2 sec2 x tan x  

Conclusion

With this, we conclude our introduction on the derivative of the function. References