Unformatted text preview: Implicit differentation
Consider the equation The graph is below. This equation is not a function. A point on the curve is . Does the curve have slope there? Does it have a tangent line there? If we just look at the graph "near" , it is a function - see below If we can treat the variable as a function, i.e, as derivative from the equation.
MTH 173 notes section 3.6, page 1 , then we can "find" the But we've seen that the chain rule tells us how to differentiate functions of functions. For example, we know that the derivative of with respect to is or , and we , know that the derivative of is , or, equivalently, given the derivative is - in short, if we consider a function of , then the derivative of with respect to is So, we have If we consider the wrt , we obtain to be an unknown function of and differentiate this equation which we can solve for Then, at the point on the curve, we have this is the slope at the point The equation of the tangent line is . If we calculate the derivative at this point, Note that another point on the graph is we obtain the tangent at this point is vertical. The graph below shows that behavior. MTH 173 notes section 3.6, page 2 Implicit differentiation is the process we use when we have an equation in and and we are to determine . We treat as a and differentiate using the chain rule, as appropriate. differentiate wrt 6. 8. 16. left side: MTH 173 notes section 3.6, page 3 26. eqn of tangent to at equation is: We can use implicit differentation to discover rules for the derivatives of the inverse trig functions. Remember that, within the appropriate restriction, we learned MTH 173 notes section 3.6, page 4 arcsin differentiate cos sin sin to obtain cos sin , we can replace cos with sin , But since we know that: cos giving sin and then since sin , we have arcsin , we have the rule for But now since orginally we had that the derivative of this function: arcsin Similarly we can obtain differentiation rules for the remaining inverse trig functions. They are listed on page 233. 42. arcsin arcsin arcsin MTH 173 notes section 3.6, page 5 60. show orthogonal: (1) (2) the two circles intersect on the line the points of intersection: I replace with in the equation to give: then substitute these in to obtain intersection points: At the origin, the tangent lines are the and axes. and : At the other point, we multiply the derivatives (1) and (2) after substituting for , so that the curves are orthogonal at this point of intersection. MTH 173 notes section 3.6, page 6 ...
View Full Document