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 quotient rule states that the derivative of a function h(x) where h(x) = f(x)/g(x) is h'(x) = (g(x)f'(x) - f(x)g'(x))/(g(x))^{2}.
For example, let h(x)=x^{2}/4x^{3}-7. Our functions f and g are f(x)=x^{2} and g(x)=4x^{3}-7. By using normal differentiating rules, we know that f'(x)=2x and g'(x)=12x^{2}. Plug these values into the formula: h'(x)=((4x^{3}-7)(2x) - (x^{2})(12x^{2}))/(12x^{2})^{2}. By expanding, we get h'(x)=(8x^{4}-14x - 12x^{4})/144x^{4} = (-4x^{4}-14x)/144x^{4}. Simplified, this becomes h'(x)=(-2x^{3}-7)/72x^{3}.
Let's do a harder problem. Now sometimes you're going to be faced with something like this where, you're not actually given a formula for a function. Instead, you're going to be given data.
For example I have two functions here v, and w. I'm told that v(3) is 10. V'(3) is 6. W(3) is 5. W'(3) is -7. I want to find h(3), and h'(3). These are values for a completely different function, but a function that's defined as the quotient of v(x) over w(x).
So first thing is let's find h(3). That should be pretty easy. Remember that h(x) according to this, is v(x) over w(x). So h(3) will be v(3) over w(3). H(3) is v(3) over w(3). That's going to be 10 over 5 which is 2. So that's one answer. H(3) is 2.
So now I have to find h'(3). First I want to represent h'(x). So remember h(x) is v(x) over w(x). So we have a high function, and a low function. So h'(x) is low d high. So w(x), v'(x) minus high d low v(x), w'(x) over the square of what's below. Again the denominator is w, so [w(x)]².
Now I just have to pop in 3. So h'(3) is w(3) times v'(3) minus v(3), w'(3.) over [w(3)]². Then w(3), 5. V'(3), 6. So 5 times 6 for these two minus v(3), w'(3). So that's 10 times -7 all over [w(3)]². W(3) again, 5. So this is going to be 25. 5² is 25. In the numerator, I've got 30, minus, minus, plus 70, a 100 over 25. That's 4. So that gives me my second answer, h'(3) is 4.
Now part be asks me to use this information, to find an equation of the line tangent to, it should be y equals h(x) at x equals 3. Now when you're finding the equation of the tangent line, you need two things. You need the point of tangency, both coordinates, and you need the slope.
Now one of the coordinates is going to be x equals 3. The second coordinate is h(3) which we already found to be 2. So the point of tangency is 3,2. What about the slope? Well, the slope is h'(3) which is 4. So the slope, 4.
Then we just use point-slope. Remember point slope is y minus y1 equals m times x minus x1. X1, and y1 are the point 3, and 2. So we have y minus the y coordinate 2 equals m which is 4, times x minus the x coordinate which is 3. X minus 3. That's the equation for your tangent line.
Unit
Techniques of Differentiation