So, basically following the same method, I have made a mistake this is not just summation it is log base 2 log base 2 of and summation I equal to 1 through L please note that if the capacity is not just the quantity within bracket is logarithm base 2 that is what gives us the capacity. So, now I would have to differentiate each of these terms with respect to P i and notice when I differentiate with respect to P i I will get only one term from the first expression and one term from the second expression and also keeping in mind that we can always write logarithm of a quantity to the base 2 can be written in terms of the logarithm of the quantity base e multiplied by logarithm of e base 2. So, basically I would like to write it in terms of logarithm natural logarithm times log e base 2. So, this will give me log e base 2 that is a constant and when I differentiate the natural logarithm of 1 plus sigma i squared rho P i then I get 1 plus sigma I squared rho P i and then in the numerator sigma i squared times P rho sigma I times rho I am differentiating with respect to P i then on the other side we get minus lambda equal to 0. So, this is a similar expressions we will get for i equal to 1 all the way through L.
अब मुझे Pi के संबंध में इन सभी टर्म्स को डिफरेंशियेट करना होगा और तब मुझे पहली एक्सप्रेशन में से केवल एक टर्म मिलेगा और दूसरी एक्सप्रेशन से एक टर्म मिलेगा, यह ध्यान में रखे कि होता है, इसलिए हमें मिलता है। तो, हमें i = 1 से L तक यह समान एक्सप्रेशन मिलेगा।