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# Business economics basics of math derivatives

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### Business economics basics of math derivatives

1. 1. Business Economics – Basics of Math Derivatives 7 th Oct 2009 Sameer Gunjal
2. 2. Derivatives <ul><li>In calculus (a branch of mathematics) the derivative is a measure of how a function changes as its inputs change. </li></ul>
3. 3. Notations used <ul><li>Leibniz's notation </li></ul><ul><ul><li>dy/dx, d 2 y/dx 2 </li></ul></ul><ul><li>Lagrange's notation </li></ul><ul><ul><li>f’(x), f’’(x) </li></ul></ul>
4. 4. Derivatives of powers <ul><li>If, f(x) = x n </li></ul><ul><li>Then, first derivative would be </li></ul><ul><ul><li>F’(x) = n . x (n-1) </li></ul></ul><ul><li>Second derivative would be </li></ul><ul><ul><li>F’’(x) = n . (n-1) . x (n-2) </li></ul></ul>
5. 5. Derivatives of constant <ul><li>Derivative of a constant is ‘0’. </li></ul><ul><li>Eg. </li></ul><ul><ul><li>f(x) = a </li></ul></ul><ul><ul><li>f’(x) = 0 </li></ul></ul><ul><ul><li>Ex. 2 </li></ul></ul><ul><ul><li>f(x) = a + x n </li></ul></ul><ul><ul><li>f‘(x) = 0 + n . x (n-1) </li></ul></ul>
6. 6. Maxima and Minima <ul><li>Maxima and minima, the highest and lowest values (points) of a function in calculus </li></ul><ul><li>f(x) = x 2 </li></ul><ul><li>f(x) = x (1/x) </li></ul>
7. 7. Test to identify Maxima and Minima <ul><li>In calculus, the first derivative test determines whether a given critical point of a function is a maximum or a minimum </li></ul><ul><ul><li>For a given function f(x), find f’(x). </li></ul></ul><ul><li>The second derivative test is a criterion often useful for determining whether a given stationary point of a function is a local maximum or a local minimum </li></ul><ul><ul><li>For the above function differentiate twice to get f’’(x), </li></ul></ul><ul><ul><ul><li>If f’’(x) < 0 then has a local maximum at . </li></ul></ul></ul><ul><ul><ul><li>If f’’(x) > 0 then has a local minimum at . </li></ul></ul></ul>
8. 8. Steps to be followed for solving optimization problems <ul><li>Step 1 : Identify the function to be optimized. </li></ul><ul><li>Step 2 : Bring the function in one variable. </li></ul><ul><li>Step 3 : Differentiate the function w.r.t. to the variable and equate it to 0 to find a value of the variable </li></ul><ul><ul><li>f’(x) = 0  Find one value of ‘x’ </li></ul></ul><ul><li>Step 4 : Differentiate f’(x) w.r.t. to the variable. </li></ul><ul><ul><li>Find f’’(x) and substitute the above value of x in it. </li></ul></ul><ul><li>Step 5 : </li></ul><ul><ul><li>If the value is > 0 then it is a ‘minima’ </li></ul></ul><ul><ul><li>If the value is < 0 then it is a ‘maxima’ </li></ul></ul>
9. 9. Example - 7 <ul><li>Mr. Mike has estimated the demand function for CDs sold by his firm ABC Limited on a monthly basis. </li></ul><ul><li>Qd = 1200 – 20Pd + 0.1*I + 0.08*A </li></ul><ul><li>where, </li></ul><ul><ul><li>Qd = Quantity demanded </li></ul></ul><ul><ul><li>Pd = Price per CD </li></ul></ul><ul><ul><li>I = Per capita income in the market </li></ul></ul><ul><ul><li>A = Advertising expenditure </li></ul></ul><ul><li>  </li></ul><ul><li>Currently the per capital income is Rs. 1200 per month. The firm expends around Rs. 4000 per month on advertising. Mike wants to maximize his monthly sales revenues to fulfil this objective what price should he charge for the CDs and what will be the monthly revenue if he achieved maximum monthly sales. </li></ul>
10. 10. <ul><li> Thank You </li></ul>