What Is Second Order Derivative?

Before knowing what is second-order derivative, let us first know what a derivative means. Basically, a derivative provides you with the slope of a function at any point. Now, what is a second-order derivative? A second-order derivative is a derivative of the derivative of a function. It is drawn from the first-order derivative.  So we first find the derivative of a function and then draw out the derivative of the first derivative. A first-order derivative can be written as f’(x) or dy/dx whereas the second-order derivative can be written as f’’(x) or d²y/dx² 

A second-order derivative can be used to determine the concavity and inflexion points. 

Concavity

Concave up: The second derivative of a function is said to be concave up or simply concave, at a point (c,f(c)) if the derivative  (d²f/dx²)_x=c >0. In such a case, the points of the function neighbouring c will lie above the straight line on the graph which will be tangent at the point (c, f(c)). Here is a figure to help you to understand better. 

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Concave Down: Concave down or simply convex is said to be the function if the derivative (d²f/dx²)_x=c at a point (c,f(c)). In such a case, the points of the function neighbouring c will lie below the straight line on the graph which is tangent at the point (c,f(c)). Here is a figure to help you to understand better. 

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Point of Inflection

The point of inflexion can be described as a point on the graph of the function where the graph changes from either concave up to concave down or concave down to concave up. The sigh of the second-order derivative at this point is also changed from positive to negative or from negative to positive. The second-order derivative of the function is also considered 0 at this point. 

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Second-Order Derivative Examples

Question 1) If f(x) = sin3x cos4x, find  f’’(x). Hence, show that,  f’’(π/2) = 25.

Solution 1) We have, 

f(x) =  sin3x cos4x or, f(x) = \[\frac{1}{2}\] . 2sin3x cos4x = \[\frac{1}{2}\](sin7x-sinx) 

Differentiating two times successively w.r.t. x we get, 

f’(x) = \[\frac{1}{2}\] [cos7x . \[\frac{d}{dx}\]7x-cosx] = \[\frac{1}{2}\] [7cos7x-cosx]

And f’’(x) = \[\frac{1}{2}\] [7(-sin7x)\[\frac{d}{dx}\]7x-(-sinx)] = \[\frac{1}{2}\] [-49sin7x+sinx]

Therefore,f’’(π/2) = \[\frac{1}{2}\] [-49sin(7 . π/2)+sin π/2] = \[\frac{1}{2}\] [-49 . (-1)+1]

[sin7 . π/2 = sin(7.π/2+0) = – cos0= -1]

                   = \[\frac{1}{2}\] x 50 = 25(Proved)

Question 2) If y = \[tan^{-1}\] (\[\frac{x}{a}\]), find y₂.

Solution 2) We have,  y = \[tan^{-1}\] (\[\frac{x}{a}\])

Differentiating two times successively w.r.t. x we get,

y₁ = \[\frac{d}{dx}\] (\[tan^{-1}\] (\[\frac{x}{a}\])) = \[\frac{1}{1+x²/a²}\] . \[\frac{d}{dx}\](\[\frac{x}{a}\]) = \[\frac{a²}{x²+a²}\] . \[\frac{1}{a}\] = \[\frac{a}{x²+a²}\]

And, y₂ = \[\frac{d}{dx}\] \[\frac{a}{x²+a²}\] = a . \[\frac{d}{dx}\] (x²+a²)^-1 = a . (-1)(x²+a²)^-2 .  \[\frac{d}{dx}\] (x²+a²)  

             = \[\frac{-a}{ (x²+a²)²}\] . 2x = \[\frac{-2ax}{ (x²+a²)²}\] 

Question 3) If y = \[e^{2x}\] sin3x,find y’’. 

Solution 3) We have, y = \[e^{2x}\]sin3x 

Differentiating two times successively w.r.t. x we get,

y’ = \[\frac{d}{dx}\](\[e^{2x}\]sin3x) = \[e^{2x}\] . \[\frac{d}{dx}\]sin3x + sin3x .  \[\frac{d}{dx}\] \[e^{2x}\]

Or,   

y’ = \[e^{2x}\] . (cos3x) . 3 + sin3x . \[e^{2x}\] . 2 = \[e^{2x}\] (3cos3x + 2sin3x)

And    

y’’ = \[e^{2x}\]\[\frac{d}{dx}\](3cos3x + 2sin3x) + (3cos3x + 2sin3x)\[\frac{d}{dx}\] \[e^{2x}\]

        = \[e^{2x}\][3.(-sin3x) . 3 + 2(cos3x) . 3] + (3cos3x + 2sin3x) . \[e^{2x}\] . 2  

        = \[e^{2x}\](-9sin3x + 6cos3x + 6cos3x + 4sin3x) =  \[e^{2x}\](12cos3x – 5sin3x)

Question 4) If y = acos(log x) + bsin(log x), show that,

                     x²\[\frac{d²y}{dx²}\] + x \[\frac{dy}{dx}\] + y = 0

Solution 4) We have, y = a cos(log x) + b sin(log x)

Differentiating both sides of (1) w.r.t. x we get,

\[\frac{dy}{dx}\] = – a sin(log x) . \[\frac{1}{x}\] + b cos(log x) . \[\frac{1}{x}\]

Or, 

x\[\frac{dy}{dx}\] = -a sin (log x) + b cos(log x)

Differentiating both sides of (2) w.r.t. x we get,

x . \[\frac{d²y}{dx²}\] +  \[\frac{dy}{dx}\] . 1 = – a cos(log x) . \[\frac{1}{x}\] – b sin(log x) . \[\frac{1}{x}\]

Or,     

x²\[\frac{d²y}{dx²}\] + x\[\frac{dy}{dx}\] = -[a cos(log x) + b sin(log x)]

 Or,  

x²\[\frac{d²y}{dx²}\] + x\[\frac{dy}{dx}\] = -y[using(1)]     

Or, 

x²\[\frac{d²y}{dx²}\] + x\[\frac{dy}{dx}\] + y = 0 (Proved)   

Question 5) If y = \[\frac{1}{1+x+x²+x³}\], then find the values of

[\[\frac{dy}{dx}\]]x = 0 and [\[\frac{d²y}{dx²}\]]x = 0

Solution 5) We have, y = \[\frac{1}{1+x+x²+x³}\] 

Or,  

y =   \[\frac{x-1}{(x-1)(x³+x²+x+1}\] [assuming x ≠ 1]    

    = \[\frac{x-1}{(x⁴-1)}\]       

Differentiating two times successively w.r.t. x we get,

\[\frac{dy}{dx}\] = \[\frac{(x⁴-1).1-(x-1).4x³}{(x⁴-1)²}\] = \[\frac{(-3x⁴+4x³-1)}{(x⁴-1)²}\]…..(1)

And,  

\[\frac{d²y}{dx²}\] = \[\frac{(x⁴-1)²(-12x³+12x²)-(-3x⁴+4x³-1)2(x⁴-1).4x³}{(x⁴-1)⁴}\]…..(2)

Putting x = 0 in (1) and (2) we get, 

[\[\frac{dy}{dx}\]] x = 0 = \[\frac{-1}{(-1)²}\] = 1 and [\[\frac{d²y}{dx²}\]] x = 0 = \[\frac{(-1)².0 – 0}{(-1)⁴}\] = 0