Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
The sum rule of differentiation states that the derivative of a sum is the sum of the derivatives. So, when finding the derivative of a polynomial function, you can look at each term separately, then add the results to find the derivative of the entire function. For example, let f(x)=x^{3}+2x+5. Now look at these derivatives separately. By the power rule, the derivative of x^{3} is 3x^{2}, the derivative of 2x is 2, and the derivative of 5 is 0 (the derivative of a constant is 0). So, the derivative is f'(x)=3x^{2}+2+0=3x^{2}+2.
Let's do a problem. I want to differentiate these three functions. They are not all polynomial functions, but I'll still get to use my constant multiple rule, and my sum rule of differentiation.
First let's take the derivative of this function. This is a polynomial x³ plus 2x plus 5. So the derivative of this would be dy/dx. That's the derivative with respect to x of x³ plus 2x plus 5. Now I can use the sum rule here, because x³ is a power function, and 2x plus 5 is a linear function. I know how to differentiate those. So I have the derivative of x³ plus the derivative of 2x plus 5.
So this is going to be 3x², and this is going to be 2. 3x² plus 2. Easy. Now let's take a look at this guy. First of all, recall that the square root of x is a power function that can be written as 2x to the ½. So you need the constant multiple rule here. The derivative of y; dy/dx, is the derivative with respect to x of 2x to the ½. So I pull constant outside, and I get 2 times the derivative of x to the ½.
Remember the power rule. The exponent comes out in front, and it's replaced by one less. So it's going to be 2 times ½ comes out in front, x to the ½ minus 1, -½. So that's just x to the -½, which is 1 over x to the ½, 1 over root x. That's the answer.
Then part c I have two functions here. 1 over x, remember that that can be written as x to the -1. So that really is a power function so I can use the power rule in it. So the derivative is going to be the derivative with respect to x of 3x minus x to the -1. I can use the sum rule first, because I have two terms in this function. So the derivative with respect to x of 3x plus the derivative with respect to x of -x to the -1. Technically I need to use the constant multiple rule. I have a -1 in front here. I can pull that out, and that gives me plus -1 times the derivative of x to the -1.
So now I'm ready to go. 3x this is a linear function. Its derivative is just going to be a slope, so plus-1 times, and the derivative of x to the -1 again the power rule. The exponent comes out in front. -1 x to the, then I subtract 1 from the exponent -1, minus another 1 is -2. So what I have here is 3, -1 times -1 is +1. So plus x to the -2. 3 plus 1 over x², and that's my answer.
Unit
The Derivative