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