Solution: The point on the curve y = at x = 0 is (0, = (0,1). The slope of the tangent line ex e0) is . At x = 0 the slope = = 1. Thus the equation of the tangent line is y 1 = 1(x 0) or e0 – – simplified we get y = x + 1. Find the derivative of = f(x) x2ex. Solution: the product rule gives . In most applications the function involved is not just but rather instead functions which have the form . As we will see, this type of function can be differentiated using the Chain Rule. Suppose that y = . Let = Then y = . By the Chain Rule u f(x). This result is worth emphasizing. Find the derivative of . Solution: Using the formula in the box above we get Derivative of For any positive constant 0 ≠ Find the derivative of . Solution: Using the formula above, Suppose that, The derivative will be given by Find the derivative of . Solution: Using the formula above, .
Solution: The point on the curve y = ex at x = 0 is (0, e0) = (0,1). The slope of the tangent line is . At x = 0 the slope = e0 = 1. Thus the equation of the tangent line is y – 1 = 1(x – 0) or simplified we get y = x + 1. Find the derivative of f(x) = x2ex. Solution: the product rule gives . In most applications the function involved is not just but rather instead functions which have the form . As we will see, this type of function can be differentiated using the Chain Rule. Suppose that y = . Let u = f(x). Then y = . By the Chain Rule This result is worth emphasizing. Find the derivative of . Solution: Using the formula in the box above we get Derivative of For any positive constant ≠ 0 Find the derivative of . Solution: Using the formula above, Suppose that, The derivative will be given by Find the derivative of . Solution: Using the formula above, .
BC%20Notes%20Test%202.docx