Differentiation Mathematical Economics

Differentiation Mathematical Economics

 Formulas:

Differentiation is a process of calculating a function that represents the rate of change of one variable with respect to another. Differentiation and derivatives have immense application not only in our day-to-day life but also in higher mathematics.

: white; color: #353535; font-family: Merriweather, Merriweather-Fallback, "Times New Roman", "Times, Serif"; margin: 0px 0px 1em; padding: 0px;">Differentiation Definition: Let’s say y is a function of x and is expressed as \(y=f(x)\). Then, the rate of change of “y” per unit change in “x” is given by \(\frac{dy}{dx} \).

Here, \(\frac{dy}{dx} \) is known as differentiation of y with respect to x. It is also denoted as \({f}'(x)\).

Rules Of Differentiation: Differentiation Formulas   

There are mainly 7 types of differentiation rules that are widely used to solve problems relate to differentiation:

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1. Power Rule: When we need to find the derivative of an exponential function, the power rule states that:
   \(\frac{d}{dx}{{x}^{n}}=n\times {{x}^{n-1}}\)


2. Product Rule: When \(f(x)\) is the product of two functions, \(a(x)\) and \(b(x)\), then the product rule states that:
   \(\frac{d}{dx}f(x)=\frac{d}{dx}\left[ a(x)\times b(x) \right]=b(x)\times \frac{d}{dx}a(x)+a(x)\times \frac{d}{dx}b(x)\)


3. Quotient Rule: When \(f(x)\) is of the form \(\frac{a(x)}{b(x)}\), then the quotient rule states that:
   \(\frac{d}{dx}f(x)=\frac{d}{dx}\left[ \frac{a(x)}{b(x)} \right]=\frac{b(x)\times \frac{d}{dx}a(x)-a(x)\times \frac{d}{dx}b(x)}{{{\left[ b(x) \right]}^{2}}}\)


4. Sum or Difference Rule: When a function \(f(x)\) is the sum or difference of two functions \(a(x)\) and \(b(x)\), then the sum or difference formula states that:
   \(\frac{d}{dx}f(x)=\frac{d}{dx}\left[ a(x)\pm b(x) \right]=\frac{d}{dx}a(x)\pm \frac{d}{dx}b(x)\)


5. Derivative of a Constant: Derivative of a constant is always zero.
   Suppose \(f(x)=c\), where c is a constant. We have,
   \(\frac{d}{dx}f(x)=\frac{d}{dx}(c)=0\)


6. Derivative of a Constant Multiplied with a Function \(f\): When we need to find out the derivative of a constant multiplied with a function, we apply this rule:
   \(\frac{d}{dx}\left[ c\times f(x) \right]=c\times \frac{d}{dx}f(x)\)


7. Chain Rule: The chain rule of differentiation states that:
   \(\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\)

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Differentiation Formulas List

: white; color: #353535; font-family: Merriweather, Merriweather-Fallback, "Times New Roman", "Times, Serif"; margin: 0px 0px 1em; padding: 0px;">The table below provides the derivatives of basic functions, constant, a constant multiplied with a function, power rule, sum and difference rule, product and quotient rule, etc. Differentiation formulas of basic logarithmic and polynomial functions are also provided.

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: initial; background-clip: initial; background-color: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial; background: -webkit-gradient(linear, 0% 0%, 0% 100%, from(rgb(251, 251, 251)), to(rgb(250, 250, 250))); border-bottom: 1px solid rgb(224, 224, 224); border-left: 0px; border-top: 1px solid rgb(255, 255, 255); padding: 18px 18px 18px 20px; text-align: left;">(i) \(\frac{d}{dx} (k)= 0\)

(ii) \(\frac{d}{dx} (ku)= k\frac{du}{dx}\)

(iii) \(\frac{d}{dx} (u±v)= \frac{du}{dx}±\frac{dv}{dx}\)

(iv) \(\frac{d}{dx} (uv)= u\frac{dv}{dx}+v\frac{du}{dx}\)

(v) \(\frac{d}{dx} (u/v)= \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\)

: initial; background-clip: initial; background-color: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial; background: -webkit-gradient(linear, 0% 0%, 0% 100%, from(rgb(251, 251, 251)), to(rgb(250, 250, 250))); border-bottom: 1px solid rgb(224, 224, 224); border-left: 0px; border-top: 1px solid rgb(255, 255, 255); padding: 18px 18px 18px 20px; text-align: left;">(vi) \(\frac{dy}{dx}.\frac{dx}{dy}= 1\)

(vii) \(\frac{d}{dx} (x^n)= nx^{n-1}\)

(viii) \(\frac{d}{dx} (e^x)= e^x\)

(ix) \(\frac{d}{dx} (a^x)= a^x\log a\)

(x) \(\frac{d}{dx} (\log x)= \frac{1}{x}\)

(xi) \(\frac{d}{dx} \displaystyle \log _{a}x= \frac{1}{x}\displaystyle \log _{a}e\)

(xii) \(\frac{d^n}{dx^n} (ax+b)^n= n!a^n\)

Let us now look into the differentiation formulas for different types of functions.

Differentiation Formulas For Trigonometric Functions

Sine (sin), cosine (cos), tangent (tan), secant (sec), cosecant (cosec), and cotangent (cot) are the six commonly used trigonometric functions each of which represents the ratio of two sides of a triangle. The derivatives of trigonometric functions are as under:

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(i) \(\frac{d}{dx} (\sin x)= \cos x\)

(ii) \(\frac{d}{dx} (\cos x)= -\sin x\)

(iii) \(\frac{d}{dx} (\tan x)= \sec^2 x\)

(iv) \(\frac{d}{dx} (\cot x)= – cosec^2 x\)

(v) \(\frac{d}{dx} (\sec x)= \sec x \tan x\)

(vi) \(\frac{d}{dx} (cosec x)= – cosec x \cot x\)

(vii) \(\frac{d}{dx} (\sin u)= \cos u \frac{du}{dx}\)

(viii) \(\frac{d}{dx} (\cos u)= -\sin u \frac{du}{dx}\)

(ix) \(\frac{d}{dx} (\tan u)= \sec^2 u \frac{du}{dx}\)

(x) \(\frac{d}{dx} (\cot u)= – cosec^2 u \frac{du}{dx}\)

(xi) \(\frac{d}{dx} (\sec u)= \sec u \tan u \frac{du}{dx}\)

(xii) \(\frac{d}{dx} (cosec u)= – cosec u \cot u \frac{du}{dx}\)

Differentiation Formulas For Inverse Trigonometric Functions

Inverse trigonometric functions like (\(\sin^{-1}~ x)\) , (\(\cos^{-1}~ x)\) , and (\(\tan^{-1}~ x)\) represents the unknown measure of an angle (of a right angled triangle) when lengths of the two sides are known. The derivatives of inverse trigonometric functions are as under:

(i) \(\frac{d}{dx}(\sin^{-1}~ x)\) = \(\frac{1}{\sqrt{1-x^2}}\)

(ii) \(\frac{d}{dx}(\cos^{-1}~ x)\) = -\(\frac{1}{\sqrt{1-x^2}}\)

(iii) \(\frac{d}{dx}(\tan^{-1}~ x)\) = \(\frac{1}{{1+x^2}}\)

(iv) \(\frac{d}{dx}(\cot^{-1}~ x)\) = -\(\frac{1}{{1+x^2}}\)

(v) \(\frac{d}{dx}(\sec^{-1}~ x)\) = \(\frac{1}{x\sqrt{x^2-1}}\)

(vi) \(\frac{d}{dx}(coses^{-1}~ x)\) = -\(\frac{1}{x\sqrt{x^2-1}}\)

(vii) \(\frac{d}{dx}(\sin^{-1}~ u)\) = \(\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}\)

(viii) \(\frac{d}{dx}(\cos^{-1}~ u)\) = -\(\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}\)

(ix) \(\frac{d}{dx}(\tan^{-1}~ u)\) = \(\frac{1}{{1+u^2}}\frac{du}{dx}\)

Differentiation Formulas For Hyperbolic Functions

The hyperbolic function of an angle is expressed as a relationship between the distances from a point on a hyperbola to the origin and to the coordinate axes. The derivatives of hyperbolic functions are as under:

(i) \(\frac{d}{dx} (\sinh~ x)= \cosh x\)

(ii) \(\frac{d}{dx} (\cosh~ x) = \sinh x\)

(iii) \(\frac{d}{dx} (\tanh ~x)= sech^{2} x\)

(iv) \(\frac{d}{dx} (\coth~ x)=-cosech^{2} x\)

(v) \(\frac{d}{dx} (sech~ x)= -sech x \tanh x\)

(vi) \(\frac{d}{dx} (cosech~ x ) = -cosech x \coth x\)

(vii) \(\frac{d}{dx}(\sinh^{-1} ~ x)\) = \(\frac{1}{\sqrt{x^2+1}}\)

(viii) \(\frac{d}{dx}(\cosh^{-1} ~ x)\) = \(\frac{1}{\sqrt{x^2-1}}\)

(ix) \(\frac{d}{dx}(\tanh^{-1} ~ x)\) = \(\frac{1}{{1-x^2}}\)

(x) \(\frac{d}{dx}(\coth^{-1} ~ x)\) = -\(\frac{1}{{1-x^2}}\)

(xi) \(\frac{d}{dx}(\sec h^{-1} ~ x)\) = -\(\frac{1}{x\sqrt{1-x^2}}\)

(xii) \(\frac{d}{dx}(cos h^{-1} ~ x)\) = -\(\frac{1}{x\sqrt{1+x^2}}\) derivatives of trigonometric functions? The commonly used derivatives of six trigonometric functions are: (d/dx) sin x = cos x (d/dx) cos x = -sin x (d/dx) tan x = sec^2 x (d/dx) sec x = sec x tan x (d/dx) cot x = -cosec^2 x (d/dx) cosec x = -cosec x cot x d/dx d/dx is the general representation of the derivative function. d/dx denotes the differentiation with respect to the variable x. UV formula? (d/dx)(uv) = v(du/dx) + u(dv/dx) This formula is used to find the derivative of the product of two functions. Example of differentiation The example of differentiation is velocity which is equal to rate of change of displacement with respect to time. Another example is acceleration which is equal to rate of change of velocity with respect to time. Examples If y = x4, dy/dx = 4x3 If y = 2x4, dy/dx = 8x3 If y = x5 + 2x-3, dy/dx = 5x4 - 6x-4 Example Find the derivative of: divide each term on the numerator by 3x½. We get: (1/3)x3/2 + (5/3)x½ - x-½ (using the laws of indices). So differentiating term by term: ½ x½ + (5/6)x-½ + ½x-3/2. Notation There are a number of ways of writing the derivative. They are all essentially the same: (1) If y = x2, dy/dx = 2x This means that if y = x2, the derivative of y, with respect to x is 2x. (2) d (x2) = 2x      dx This says that the derivative of x2 with respect to x is 2x. (3) If f(x) = x2, f'(x) = 2x This says that is f(x) = x2, the derivative of f(x) is 2x. Example What is the gradient of the curve y = 2x3 at the point (3,54)? dy/dx = 6x2 When x = 3, dy/dx = 6× 9 = 54 Differentiate x5 with respect to x. Solution: Given, y = x5 On differentiate 5x Some important question Solution : f(x) = x - 3 sinx f'(x)  =  1 - 3 cos x Question 2 : Differentiate y = sin x + cos x Solution : f(x) = sin x + cos x f'(x)  =  cos x - sin x Question 3 : Differentiate f(x) = x sin x Solution : f(x) = x sin x We have to use the product rule to find the derivative. u  =  x ==>  u'  =  1 v  =  sin x ==>  v'  =  cos x Product rule : d(uv)  =  uv' + vu' f'(x)  =  x(cos x) + sin x (1) f'(x)  =  x cos x + sin x Question 4 : Differentiate y = cos x - 2 tan x Solution : f(x)  =  cos x - 2 tan x f'(x)  =  -sin x - 2 sec2 x Question 5 : Differentiate g(t) = t3cos t Solution : We have to use the product rule to find the derivative. u  =  t3 ==>  u'  =  3t2 v  =  cos t ==>  v'  =  -sin t f('x)  =  t3(-sin t) + cos t (3t2) f('x)  =  -t3sin t + 3t2cos t =  t2 (3 cos t - t sin t) Question 6 : Differentiate g(t) = 4 sec t + tan t Solution : g(t) = 4 sec t + tan t g'(t)  =  4 sec t tan t + sec2 t Question 7 : Differentiate y = ex sin x Solution : y = ex sin x u = ex  ===> u'  =  ex  v = sin x ===> v'  =  cos x  y'  =  ex (cos x) + sin x(ex) y' =  ex (cos x + sin x) Question 8 : Differentiate y  =  tan x / x Solution : y  =  tan x / x u = tan x  ===> u'  =  sec2 x  v = x ===> v'  =  1 Quotient rule : d(u/v)  =  (vu' - uv') / v2 dy/dx  =  (x sec2 x  - tan x (1)) / x2 =  (x sec2 x  - tan x) / x2  Question 9 : Differentiate y  =  sin x / (1 + cos x) Solution : y  =  sin x / (1 + cos x) u = sin x ===> u'  =  cos x  v = (1 + cos x) ===> v'  =  - sin x Quotient rule : d(u/v)  =  (vu' - uv') / v2 dy/dx  =  ((1 + cos x) cos x - sin x (-sin x)) / (1 + cos x)2 dy/dx  =  (cos x + cos2 x + sin2 x) / (1 + cos x)2 dy/dx  =  (1 + cos x) / (1 + cos x)2 dy/dx  =  1/(1 + cos x) Question 10 : Differentiate y  =  x / (sin x + cos x) Solution : y  =  x / (sin x + cos x) u = x ===> u'  =  1  v = (sin x + cos x) ===> v'  =  cos x - sin x dy/dx  =  [(sinx+cosx) (1)-x(cosx-sinx)]/(sin x+cosx)2 dy/dx  =  [sinx + cosx - x cosx + xsinx)]/(sin x+cosx)2 Students this are basic concept in my next blog I will give you detail concept which we use in Mathematical Economics For further details iemsnet.com