# Second-Order Derivative

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.

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.

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.

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

Q1. Is the Second-order Derivatives an Acceleration?

Ans. When we move fast, the speed increases and thus with the acceleration of the speed, the first-order derivative also changes over time.

Therefore we use the second-order derivative to calculate the increase in the speed and we can say that acceleration is the second-order derivative.

Q2. What do we Learn from Second-order Derivatives?

Ans. Second order derivatives tell us that the function can either be concave up or concave down. It also teaches us:

1. When the 2nd order derivative of a function is positive, the function will be concave up.

2. When the 2nd order derivative of a function is negative, the function will be concave down.

3. If the 2nd order derivative of a function tends to be 0, then the function can either be concave up or concave down or even might keep shifting.