Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
By the product rule, the derivative of a product of functions f(x)g(x) is f(x)g'(x) + f'(x)g(x). So, in order to calculate the derivative of a product of functions, calculate the derivatives of the individual functions. Then, simply multiply the derivative of the second function to the first function, and add the derivative of the first function multiplied by the second function.
For example, if your function h(x)=x^{2}ln(x), let f(x)=x^{2} and g(x)=ln(x). We know by the rules for deriving power functions and logarithmic functions that f'(x)=2x and g'(x)=1/x. h'(x)=f(x)g'(x) + f'(x)g(x). So, h'(x)=x^{2}(1/x) + 2xln(x) = x + 2xln(x).
By using finding the derivative of this function f(x)g(x), you can find the slope of the tangent line at any given x on the graph. Then, by using the point at which you found the tangent line, find the equation of the tangent line by using point-slope form: y-y_{1}=m(x-x_{1}), where m is the slope at that point, and (x_{1},y_{1}) is the of tangency (the point where the line is tangent to the graph).
Let's do one more problem. A little harder this time. Given v(3) equals 5, v'(3) equals 2, w(3) equals 16, w'(3) equals -4. I want to (a), find h(3) and h'(3) where h(x) is the product of v and w. Then I'll find an equation of the line tangent to y equals h(x) at x equals 3.
So first h(3), based on this formula will be v(3) times w(3). Now just going back here, v(3) is 5, w(3) is 16, so this is 5 times 16. That's 80. What about h'(3). Well, using the product rule, it will be the first times the derivative of the second plus the second times the derivative of the first. So v(3) times the derivative of w(3) plus w(3) times v'(3).
So going back to our values here, v(5) is 5 times -4 plus 16 times 2. 5 times -4 plus 16 times 2. That's -20 plus 32 which is 12. That's all we were asked to find h(3) is 80. H'(3) is 12.
In part b, we are asked to find an equation of the line tangent to y equals h(x) at x equals 3. Well, it turns out we have almost everything we need. We do have everything we need. We need the point of tangency, and the slope of the tangent.
Now the point of tangency is, point x equals 3, and y is h(3) 80. The slope, well that's just h'(3) which we found to be 12. So using the point-slope formula, y minus 80 equals the slope 12 times x minus 3. That's the point-slope equation for the tangent line.
Now if you want to write it in slope-intercept form, it will be 12x minus 36. I add 80 to that, so plus 44. Y equals 12x plus 44. That's the equation of the line tangent to y equals h(x) at x equals 3.
Unit
Techniques of Differentiation