The Product Rule - Problem 3
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.