Quotient Rule for Derivatives. Example Problem
Let us now take the derivative of 3x-5 divided by 2x+1. This is a quotient rule problem because we have a fraction. We have a rational expression, and the variable is in each part of that rational expression. Please remember that if you were to take the derivative of x cubed, the quotient rule is not necessary because we are not taking a quotient of two variables. You would rewrite that expression as x to the negative 3 power, apply the power rule, and move on. All right, let’s return to the problem at hand. I feel it is important to point out that many of my students make this mistake. Remember that we call the top part hi, call the bottom part Lo. It is our mnemonic Lo times the derivative of hi minus high dLo all over Lo squared. Then you fill in the missing parts of the puzzle. Right down Lo. The result is 2x plus 1 times the derivative of hi, which equals 3x minus 5, plus the derivative of a constant multiplied by the Lo function. The derivative of 2 times x plus 1 is 2 divided by 2x plus 1 squared. Now a little cleaning up. It works good, see? We can distribute the 3, 2, and negative throughout the first parentheses. We then have 6x plus 3 minus 6x plus 10 divided by 2x plus 1 squared. You can see that 6x minus 6x is equal to zero, which means that 3 plus 10 is equal to 13 divided by 2x plus 1 squared. This is your derivative. d/dx[(3x-5)/(2x+1)] (Lo dhi-hi dLo)/Lo^2 ((2x+1)*3-(3x-5)*2)/(2x+1)^2 (6x+3-6x+10)/(2x+1)^2=13/(2x+1)^2
