SecondOrder Derivative
What Is Second Order Derivative?
Before knowing what is secondorder 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 secondorder derivative? A secondorder derivative is a derivative of the derivative of a function. It is drawn from the firstorder derivative. So we first find the derivative of a function and then draw out the derivative of the first derivative. A firstorder derivative can be written as f’(x) or dy/dx whereas the secondorder derivative can be written as f’’(x) or d²y/dx²
A secondorder 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 secondorder derivative at this point is also changed from positive to negative or from negative to positive. The secondorder derivative of the function is also considered 0 at this point.
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SecondOrder 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 = \[\frac{1}{2}\](sin7xsinx)
Differentiating two times successively w.r.t. x we get,
f’(x) = \[\frac{1}{2}\] [cos7x . \[\frac{d}{dx}\]7xcosx] = \[\frac{1}{2}\] [7cos7xcosx]
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 = \[\frac{x1}{(x1)(x³+x²+x+1}\] [assuming x ≠ 1]
= \[\frac{x1}{(x⁴1)}\]
Differentiating two times successively w.r.t. x we get,
\[\frac{dy}{dx}\] = \[\frac{(x⁴1).1(x1).4x³}{(x⁴1)²}\] = \[\frac{(3x⁴+4x³1)}{(x⁴1)²}\]…..(1)
And,
\[\frac{d²y}{dx²}\] = \[\frac{(x⁴1)²(12x³+12x²)(3x⁴+4x³1)2(x⁴1).4x³}{(x⁴1)⁴}\]…..(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 Secondorder Derivatives an Acceleration?
Ans. When we move fast, the speed increases and thus with the acceleration of the speed, the firstorder derivative also changes over time.
Therefore we use the secondorder derivative to calculate the increase in the speed and we can say that acceleration is the secondorder derivative.
Q2. What do we Learn from Secondorder Derivatives?
Ans. Second order derivatives tell us that the function can either be concave up or concave down. It also teaches us:

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

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

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.