Math for Non-Geeks/ Exercises: Derivatives 4

oben Criterion for constancy Simple application Let and . For the function we assume for all . Show that then, is constant. By assumption there is Further there is , since . The squeeze theorem then yields for all . By the computation rule for limits we then get a zero differential quotient for all . Hence is constant. Proof of identities Show that for all Part 1: The function is differentiable by the chain- and difference rule with So is constant. Since further there is we have . Part 2: is differentiable according to the sum rule, since the arcus-functions are differentiable. Further there is Hence is constant. Further there is since and . So establishing the assertion. Logarithm representations of and Show that for for Part 1: The function is differentiable, see examples for derivatives, with By the chain- and sum rule also is differentiable with So we get . But now, since , and there is So , and hence . Part 2: is differentiable, as well, with By the factor-, chain- and quotient rule, also is differentiable with So we have . Since by , as well as there is again , and hence . Extension of the identity theorem Let be twice differentiable with . Then and differ only by a linear function with . Because according to the identity theorem there is a with if we now set , then there is the identity theorem again provides us with a such that General solution of a differential equation Let be differentiable and . Further let Show that: and satisfy the differential equations. If two functions and satisfy the differential equations, then there is and . Furthermore, if and there is and , then and . Hint for part 2: Consider and for functions and that satisfy the differential equations. Part 1: There is and So and satisfy the differential equations. Part 2:We define (as given in the hint) the auxiliary functions These are differentiable by the product-, sum- and difference rule with and by the criterion for constancy, there is now with . Further there is So and . Part 3: If further and then Since and analogously both the functions and satisfy the differential equation (or ). Monotonicity criterion Monotonicity of the exponential function Show using the monotonicity criterion that: For all there is is strictly monotonically increasing. Hint: use 1. in order to prove 2. Part 1: For the differentiable auxiliary function there is So is strictly monotonically decreasing by the monotonicity criterion. Further thumb|Graph of the function and Since is continuous and strictly monotonically decreasing, there must be . Part 2: There is . Since is strictly monotonically increasing, the function is strictly monotonically increasing, if and only if the "inner function" is. This function in turn is differentiable on all of by the product rule and there is By the monotonicity criterion, the function , and hence also is strictly monotonically increasing. Condition for monotonicity of a cubic function Let . Provide a condition for such that is strictly monotonically increasing on all of . Hint: Distinguish the cases , and For being strictly monotonically increasing on all of , we need by the monotonicity criterion that holds for all . Then . For to be strictly monotonically increasing, must hold. For this is never possible for any . However, if , then there is . So is strictly monotonically increasing for and . By completing the square, we get So is strictly monotonically increasing whenever there is This is satisfied for all if and only if the right-hand side is negative. This in turn is exactly the case for Hence, is strictly monotonically increasing for and . Here, we have However, this is never fulfilled for all . So in this case is never strictly monotone increasing. Similarly, we can show that is strictly monotonically decreasing in the cases and , as well as for and . Applying the monotonicity criterion Let be differentiable with . Further let for some (fixed) and all . Show that there is for all Hint: Consider the auxiliary function . As stated in the hint we consider is differentiable according to the product rule with But now and by assumption . So there is for all by the monotonicity criterion, is monotonically decreasing. Since further there is we have for all Therefore, we also have for all Derivative and extrema Extrema of functions 1 Investigate whether the following functions have local/global extrema. Determine and characterise these if they exist. Part 1: Part 1 Local extrema of is differentiable on according to the quotient rule with According to the sufficient criterion for the existence of an extremum , there must be . Now So and are the candidates for local extrema in . Now there is The case and is not possible. Thus holds on . Further there is So holds on and on . By the sufficient criterion, is a (strict) local minimum and is a (strict) local maximum of . Part 2 Global extrema of For global extrema we first have to determine the limits and . Since and there is thumb|Graph of the function Thus is unbounded from above, and therefore has no local extremum. Further, for every power of grows slower than . Thus (As numerator and denominator are positive.) Now . Thus is a global minimum of . Part 2: Part 1 Local extrema of is differentiable on by the chain rule with thumb|Graph of the function Since now and there is, . By the necessary criterion for extrema, has no local extrema on . Since is continuous on , it follows from for all that is strictly monotonically decreasing on . Therefore has a local maximum at . Part 2 Global extrema of Using the same argument as in part 1, it follows that is even a global maximum of . Extrema of functions 2 Investigate whether the following functions are continuous, differentiability and/or have local/global extrema: Part 1 continuity and differentiability Continuity: On the function is continuous as a polynomial. The function is continuous on as a composition of the continuous functions , and . At there is In addition, since and by continuity of the exponential function So is continuous at zero and hence on all of . Differentiability: On the function is differentiable as a polynomial, with On the function is differentiable by the chain- and product rule as it is a composition of differentiable functions , and . There is At there is, according to L'Hospital's rule, So is not differentiable at zero. Part 2 Local and global extrema Local extrema: On there is . So cannot have local extrema there. On however So is a candidate for a possible extremum. Further So has a strict local minimum in . Now we still have to examine . Since is not differentiable there, our necessary and sufficient criteria are not applicable. However, there is for all and for all Thus is strictly monotonically increasing on , and strictly monotonically decreasing on . Since is continuous at zero, it follows that for all So has a strict local maximum in . Global extrema: There is thumb|Graph of the function and Therefore is unbounded from above and below, and has no global extrema. Extrema of functions 3 Show that the function has exactly two local extrema, and determine their type. Candidates for the extreme values are obtained from our necessary condition thumb|Graph of the auxiliary function Since the zeros of cannot be calculated explicitly, we need to examine this function more closely. There is , and Because of continuity and 2. with the intermediate value theorem has (at least) two zeros and . Because of 1., the function is strictly monotonically increasing on and strictly monotonically decreasing on . Thus is respectively injective on and and thus has exactly the two zeros and . For the derivative of we now have According to our first sufficient criterion, has a strict local maximum at and a strict local minimum at . Computing limits via L'Hospital L'Hospital 1 Compute the following limits with Part 1: Part 2: L'Hospital's rule is not applicable here. However, the function is continuous at the point , and therefore there is Part 3: Part 4: This limit value does not exist. First it can be decomposed into For the left-hand limit, there is now with Part 1: Analogously, however, for the right-hand limit: So , and hence does not exist. Part 5: Part 6: L'Hospital can be applied here, but it is useless: Instead, it makes sense to use the definitions of and , and then transform the quotient: Part 7: L'Hospital cannot be applied here because the enumerator for diverges (improperly). Instead, the fraction can be estimated as follows: Using the squeeze theorem, it follows . Part 8: Part 9: Part 10: L'Hospital 2 Compute the following limits: for for Part 1: and since the limit exists, the application of L'Hospital is justified. Part 2: and since the limit exists, the application of L'Hospital is justified. Part 3: and since the limit exists, the application of L'Hospital is justified. Part 4: Part 5: For the expression in the exponent there is now and since the limit exists, the application of L'Hospital is justified. By continuity of at we now get Part 6: First we have: If the limit exists, then the sequence limit also exists. Further: For the expression in the exponent there is now and since the limit exists, the application of L'Hospital is justified. By continuity of at we now get And now, we also have . Part 7: For the expression in the exponent there is now and since the limit exists, the application of L'Hospital is justified. By continuity of at we now get Part 8: For the expression in the exponent there is now and since the limit exists, the application of L'Hospital is justified. By continuity of at we now get Part 9: and since the limit exists, the application of L'Hospital is justified. Part 10: and since the limit exists, the application of L'Hospital is justified. Differentiability at a point Let Show that is continuous at zero. Show that is differentiable on and compute the derivative. Determine by 1. and 2. the derivative . Part 1: By L'Hospital there is So is continuous at zero. Part 2: Since , and are differentiable on , the quotient rule yields that is differentiable. Further, there is for : Part 3: We use the criterion from the theorem above. There is Using the criterion, is differentiable at zero with . unten