[Maths Class Notes] on Second-Order Derivative Pdf for Exam

Before knowing what a second-order derivative is, let us first know what a derivative meAns: Basically, a derivative provides you with the slope of a function at any point. The derivative of the first derivative of a function is known as the second-order derivative. The slope of the tangent at a given location, or the instantaneous rate of change of a function at that position, is determined by the first-order derivative at that point. Second-Order Derivative offers us an understanding of the shape of a function’s graph. The second derivative of the function f(x) is commonly abbreviated as f” (x). If y = f, it is sometimes expressed as D²y or y² or y” (x).

Let’s say y = f. (x)

dy/dx = f’ then (x)

If f'(x) is differentiable, we can differentiate it with respect to ‘x’ once more. The left-hand side thus becomes d/dx(dy/dx), often known as the second-order derivative of y w.r.t x.

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 inflection 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 neighboring 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 (d 2 f/dx²)x=c at a point (c,f(c)). In such a case, the points of the function neighboring 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 inflection 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

Second-Order Derivatives of a Parametric Function

We utilize the chain rule twice to determine the second derivative of the function in parametric form. To determine the second derivative, first, find the first derivative’s derivative with respect to t, then divide by the derivative of x with respect to t. If x = x(t) and y = y(t), then the second-order parametric form is:

[frac{dy}{dx}] = [frac{(dy/dt)}{(dx/dt)}] is the first derivative.

[frac{d^{2}y}{dx^{2}}] = [frac{d}{dx (dy/dx)}] is the second derivative.

[frac{(dy/dx)}{ (dx/dt)}] =[frac{ d}{dt} frac{ (dy/dx)}{ (dx/dt)}]

Note: The formula [frac{d^{2}y}{dx^{2}}] = [frac{(d^{2}y/dt^{2})}{(d^{2}x/dt^{2})}]  is completely incorrect.

The local maximum or lowest inflection point values are determined by a function’s second derivative. These can be recognised using the following criteria:

• The function f(x) has a local maximum at x if f”(x) < 0.

• The function f(x) has a local minimum at x if f”(x) > 0.

• If f”(x) = 0, it is impossible to draw any conclusions about the point x.

You can solve the following second-order differential equation:

P(x)dy/dx + Q(x)y = f d²y/dx² + P(x)dy/dx + Q(x)y = f (x)

P(x), Q(x), and f(x) are functions of x, and they are calculated using:

Inconclusive If f(x) is a polynomial, exponential, sine, cosine, or a linear mix of these, it will only work.

Parameter variation, which is a little messier but works on a broader range of functions.

However, let’s start with the scenario where f(x) = 0 (which makes it “homogeneous”):

Allow f(x) to be a differentiable function in a convenient interval. 

The graph of f(x) can then be classified as:

At the point (c, f(c), the function is said to be Concave Up, or simply Concave, if the derivative (d²f/dx²)x=c>0. The points on the graph of the function in the vicinity of c lie above the straight line that is tangent at the point (c, f(c) in this example.

At the point (c, f(c), the function is said to be Concave Down, or simply Convex, if the derivative (d²f/dx²)x=c0. The points on the graph of the function in the vicinity of c are below the straight line that is tangent at the point (c, f(c) in this case.

The function rises at the point (c, f(c) if the derivative (df/dx)x=c>0. 

You also need to know how the derivative of a function changes when x varies. The function is considered to be Concave Up if the derivative acts as an increasing function, i.e. d/dx(df/dx)>0. The function is considered to be Concave Down if the derivative acts like a declining function, i.e. d/dx(df/dx)<0. As a result, the second derivative's value is critical in establishing the shape of the function's graph.

The second-order derivative at a given position (c, f(c) is computed if f'(x) = 0 at that point. It is a Local Minimum if f”(x) > 0 at that point, and it is a Local Maximum if f”(x) < 0 at that location. Higher-order derivatives or other methods of determination are required if f"(x) = 0.