Question and Answers #91 Derivatives

Questions and Answers Sheet 91 Derivatives Question #1 How to use linear Approximation to find the value of (1.01) ? 10 Answer: Consider the function: f ( x ) = (1 + x ) with a  0 a We have: f ( 0) = 1 f  ( x ) = a (1 + x ) a −1 f  ( 0) = a Develop in the MacLaurin series until the linear term, then stop: (1 + x ) a = f ( 0) + f  ( 0) 1! Then we have for x = (1.01) 10 = 1 + 10  x + R2 ( x ) = 1 + ax + R1 ( x ) 1 and a = 10 : 100 1 + R = 1.1 + R 100 Think about the second derivative now: f  ( x ) = a ( a − 1)(1 + x ) For x0 a −2 = 90 (1 + x ) this is an increasing function, so we have: 1  1  f  ( x )  f    for 0  x  100  100  Or f  ( x )  90 (1.01) 8 So, using Lagrange's formula: Rn ( x )  x n +1 ( n + 1)! max f ( n ) ( )  ( 0, x ) we have: R  8 45 8 1.01)  0.005 2 ( 100 Question #2 How to find all critical numbers of the function f ( x ) = x 4 5 ( x − 4) 2 ? Answer: Derivative Equal to Zero: The power rule and chain rule can be used to determine the derivative. f ( x) = 14 x 2 − 72 x + 64 5x 1 5 Now we set this equal to zero: 14 x 2 − 72 x + 64 5x 1 5 =0 14 x 2 − 72 x + 64 = 0 ( ) 2 7 x2 − 36 x + 32 = 0 2 ( 7 x − 8)( x − 4 ) = 0 We get: 7x − 8 = 0 → x = 8 7 x−4 = 0 → x = 4 Derivative Not Defined: The only possible case where 14 x 2 − 72 x + 64 5x 1 5 is not defined is when the denominator is equal to zero. 1 Therefore we set 5 x 5 = 0 and solve. 1 5x 5 = 0 x=0 Therefore, the derivative is not defined at x = 0 . 0 is in the domain of f so 0 is a critical point for f . Conclusion The critical points of this function are x = 4, x = Question #3 8 ,x =0 7 x What is the antiderivative of xe ? Answer: Let y = xe x Let 𝑢 = 𝑥 ⇒ 𝑑𝑢 = 𝑑𝑥 …(i) And let, 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 (ii) ⇒ ∫ 𝑑𝑣 = ∫ 𝑒 𝑥 𝑑𝑥 ⇒ 𝑣 = 𝑒 𝑥 …(iii) ∫ 𝑥𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − ∫ 𝑒 𝑥 𝑑𝑥 ⇒ ∫ 𝑥𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − 𝑒 𝑥 + 𝐶 ⇒ ∫ 𝑥𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 (𝑥 − 1) + 𝐶 Question #4 How do I find the absolute minimum and maximum of a function using its derivatives? Answer: ( x, y ) on it. The function's gradient will be zero at the Derivating a function will give us its gradient for every minimum and maximum values. So once we've found the derivative, if we want to find the minima and maxima, we set the derivative equal to zero and solve for x . Once we've found the value of x , we should calculate point at f  ( x ) , and this will tell us if the stationary x is a minimum or maximum. We can then use a table of values to see if y is increasing or decreasing around the turning point and this will help identify if it's a local or absolute stationary point. Question #5 What does the 2nd Derivative Test tell you about the behavior of f ( x ) = x 4 ( x − 1) 3 at these critical numbers. Answer: f ( x ) = x 4 ( x − 1) 3 f  ( x ) = 4 x3 ( x − 1) + x 4  3 ( x − 1) = x3  ( x − 1)  ( 4 ( x − 1) + 3x ) = x3  ( x − 1)  ( 7 x − 4 ) 3 f  ( x ) = ( 2 2 2 ) d 3 2 2 x  ( x − 1)  ( 7 x − 4 ) + x 3  ( x − 1)  7 dx ( ) = 3x 2  ( x − 1) + x3  2 ( x − 1)  ( 7 x − 4 ) + 7 x3  ( x − 1) 2 ( = x 2  ( x − 1)  ( 3x − 3 + 2 x )  ( 7 x − 4 ) + 7 x 2 − 7 x ( ) 2 ) ( = x 2  ( x − 1)  42 x 2 − 48x + 12 = 6 x 2  ( x − 1)  7 x 2 − 8x + 2 )  4  576 f  ( 0 ) = 0 and f  (1) = 0 f    = 0  7  2401 So the Second Derivative Test says nothing about the nature of f at the crucial numbers (points) x=0,1 but assumes that the critical number (point) x = 4 yields a local minimum for f. 7 Question #6 What are some examples of non differentiable functions? Answer: There are three ways a function can be non-differentiable. We'll look at all 3 cases. Case 1 A function in non-differentiable where it is discontinuous. Example (1a) f ( x ) = cot x is non-differentiable at x = n for all integer n. x3 − 6 x 2 + 9 x Example (1b) f ( x ) = 3 is non-differentiable at 0 and at 3 and at -1 x − 2 x 2 − 3x Note that f ( x ) = x ( x − 3) 2 x ( x − 3)( x + 1) Unfortunately, the graphing utility does not show the holes at (0, -3) and (3, 0) Example 1c) Define f(x) to be 0 if x is a rational number and 1 if x is irrational. The function is non-differentiable at all x. Example 1d) description: Piecewise-defined functions my have discontiuities. Case 2 A function is non-differentiable where it has a "cusp" or a "corner point". This occurs at a if f'(x) is defined for all x near a (all x in an open interval containing a) except at a, but lim f  ( x )  lim+ f  ( x ) . (Either because they exist but are unequal or because one or both fail to exist.) x→a− Example 2a) x→a f ( x ) = x − 2 Is non-differentiable at 2. (This function can also be written: Example 2b) f ( x ) = x 2 − 4 x + 4) f ( x ) = x − 2 Is non-differentiable at 1. Case 3 A function is non-differentiable at a if it has a vertical tangent line at a. f has a vertical tangent line at a if f is continuous at a and Example 3a) f ( x ) = 2 + 3 lim f  ( x ) =  x→a x − 3 has vertical tangent line at 1. And therefore is non-differentiable at 1. Example 3b) For some functions, we only consider one-sided limts: f ( x ) = 4 − x 2 has a vertical tangent line at -2 and at 2. lim f  ( x ) Does Not Exist, but lim− f  ( x ) =  x →2 Example 3c) x→2 f ( x ) = 3 x 2 has a cusp and a vertical tangent line at 0. Question #7  1  −1 3 with respect to sin ( 3x − 4 x ) is. sec−1  2   1 − 2x  Derivative of Answer: 2 3 Explanation for correct option: Differentiate Let,  1  with respect to x. sec−1  2   1 − 2x   1  u = sec−1  2   1 − 2x  Put , x = sin ( )  = sin −1 ( x )   1    u = sec  1 − 2 ( sin ( ) )2      2 1  u = sec−1   1 − 2 ( sin ( ) ) = cos ( 2 )  cos ( 2 )   u = sec−1 ( sec ( 2 ) ) −1 (  u = 2  u = 2sin −1 ( x ) Differentiate with respect to x  du 2 = ............... ( i ) dx 1 − x 2 Differentiate sin Let, v = sin Put −1 −1 (3x − 4x ) with respect to x 3 ( 3x − 4 x ) 3 ) x = sin ( )  = sin −1 ( x )  ( ) v = sin −1 3sin ( ) − 4 ( sin ( ) )  , 3 v = sin −1 ( sin ( 3 ) ) ( 3sin ( ) − 4 ( sin ( ) ) = sin ( 3 ) 3 )  v = 3  v = 3sin −1 ( x ) Differentiate with respect to x  dv 3 = ............... ( ii ) dx 1 − x 2 Divide the equation ( i ) by ( ii )  du  2   du 2 dx  1 − x 2   =  = 3 dv 3  dv    1 − x2  dx  Derivative of 2  1  −1 3 with respect to sin ( 3x − 4 x ) is . sec−1  2  3  1 − 2x  Hence, correct answer is option (C ) . Question #8 What is the derivative of cosh ( x) ? Answer: The definition of cosh ( x ) is e x + e− x , so let's take the derivative of that: 2 d  e x + e− x    dx  2  We can bring 1 upfront. 2 1  d x d −x   e + e  2  dx dx  For the first part, we can just use the fact that the derivative of e = e : x 1  x d −x  e + e  2 dx  x For the second part, we can use the same definition, but we also have to use the chain rule. For this, we need the derivative of − x , which is simply −1 : ( 1 x e + ( −1) e − x 2 = ( 1 x −x e −e 2 ) ) e x − e− x = 2 = sinh ( x ) (definition of sin h). And that is how you are derived. Question #9 ( ) What is the derivative of f ( t ) = t + e , te − t + t ? t t 2 Answer: tet + et − 2t + 1 1 + et Answer: Solution: Explanation: Differentiation states dy dx x = t + et y = tet − t 2 + t Take the derivative of both functions x = 1 + et y = ( ) ( ) d d 2 d tet − t + (t ) dt dt dt t Use product rule for te , which states: 𝑓 ′ 𝑔 + 𝑓𝑔′ ( ) d d ( t )  et + t  et − 2t + 1 dt dt 1  et + t  et − 2t + 1 y = tet + et − 2t + 1 Thus now dy dx tet + et − 2t + 1 1 + et Question #10 How to find two different parametric equations for rectangular equation given: y = 6x - 1? Answer: There are infinitely many ways of parameterizing a curve: Solution: The simplest parametrization is to put x = t , which gives: { x=t y = 6t − 1 Another option would be 6x = t , giving: t { 6 y = t −1 x= And as a bonus why not use { 2x = 6 x = 3t y = 18t − 1 Question #11 How to find the linearization of f ( x ) = x at x=49? Answer: We have explanation: f ( x ) = and f ( x) = 1 2 x x , so f ( 49 ) = 7 , so f  ( 49 ) = 1 14 It is the tangent line y−7 = 1 ( x − 49 ) , so the linearization is 14 l ( x) = 7 + 1 ( x − 49 ) 14 Question #12 The radius of a spherical balloon is increasing at a rate of 2 centimeters per minute. How fast is the volume changing when the radius is 14 centimeters? Answer: If the radius is r, then the rate of change of r with respect to time t, d ( r ) = 2 cm/minute dt For a spherical object, volume is a function of radius r and is V (r ) = 4   r3 3 We need to find Now, But d (V ) at r = 14cm dt d d 4 4 d d (V ) =     r 3  =  3  r 2  ( r ) = 4  r 2  ( r ) dt dt  3 dt dt  3 d d ( r ) = 2cm/minute. Therefore, (V ) at r = 14 cm is: dt dt 4 142  2 cubic cm / minute = 1568   cc/minute Question #13 How to find the linearization of the function z = x y at the point (-7, 64)? Answer: The linear function that best aproximates z = −56 + 8 ( x + 7 ) − z = x y at ( −7, 64 ) is 7 7 ( y − 64 ) = 28 + 8 x − y 16 16 To get this result, we must first notice that z is a function of the two variables So, the best linear approximation x and y . Let's write z = f ( x, y ) . Lr0 ( x, y ) of f at r0 = ( x0 , y0 ) = ( −7, 64 ) is given by Lr0 ( x, y ) = f ( x0 , y0 ) + f ( x0 , y0 )  ( ( x, y ) − ( x0 , y0 ) ) Where f is the gradient of f and  is the dot product. Geometrically, this linear approxiamtion is the tangent plane of f at r0 . The gradient f serves as the derivative in the deduction of this equation, which is very similar to the deduction of the equation for the tangent line of a real function at a point. Now we need to calculate the components of the equations for the linear aproximation. value of the function at ( x0 , y0 ) : f ( x0 , y0 ) = f ( −7, 64 ) = −7  64 = −56 The gradient f ( x, y ) of f is given by the expression  f f   x f ( x, y ) =  ,  =  y , 2 y  x y   So,    7   7  f ( x0 , y0 ) =  64, −  =  8, − 16  2 64     f ( x0 , y0 ) is simply the And lastly, we have: 7  Lr0 ( x, y ) = −56 +  8, −   ( ( x, y ) − ( −7,64 ) ) = 16   7  = −56 +  8, −   ( x + 7, y − 64 ) = 16   = −56 + 8 ( x − 7 ) − 7 7 ( y − 64 ) = 28 + 8 x − y 16 16 Question #14 How to use differentials to estimate the value of cos ( 63) ? Answer: I will assume that you want to approximate ( ) rather than cosine of 63 radians (or cosine of the number cos 63 63). We'll have to switch to radian measure at some point, either before we start or while using the functions we use. Actually, we switch to numbers. Taking y = cos x , we approximate: ( ) ( ) cos 63  cos 60 + dy 3   dy = − sin   dx = − dx 2 3     dx = 63 − 60 = 3 = 3  =  180  60 (You could call the last 2 "radians", but I won't.) So dy = − 3   3 =− 2 60 120 ( )  cos ( 60 ) + dy And cos 63 cos ( 63 )  1   3 +−  2  120  More General: I believe the concepts become more obvious if we consider linear approximation, which is simply tangent line approximation. We have y = f ( x ) , and we note that f ( x ) = f ( a ) + y The linear (differential) approximation of f ( x ) near a is: f ( x )  f ( a ) + f  ( a )( x − a ) , or f ( x )  f ( a ) + f  ( a ) dx = f ( a ) + dy (This is the equation of the line tangent to the graph of the function at the point ( a, f ( a ) ) .) We are approximating  y by using the change along the tangent (which is dy ) to approximate the change on the graph of the function (which is  y ). Question #15 Find and classify all the critical points and then use the second derivative to check your results given h ( t ) = −4.9t 2 + 39.2t + 2 Answer: h ( t ) = ( −4.9 )  2  t + 39.2 h ( t ) = 0 means −9.8t + 39.2 = 0 , only critical point t=4 Now that t=4, let's compute the second derivative. However, the second derivative is 9.8 everywhere, and in particular, it is 9.8 at time t=4. The function reaches its maximum at the crucial point because the second derivative there is negative. To examine the sign of the second derivative in such places, first locate the key points where h'(t)=0. Question #16 Find the derivative of 1 x Answer: Discover the derivative of Let y = 1 x Since, 1 1 − = ( x) 2 x y = ( x) − 1 2 1 x  −1  dx 2  1 d ( y)  = − 1 x − 2 −1  =  ( ) dx dx 2 n  d (x )   = n.x n −1  dx   1 − 32 =− x 2 1 =− 3 2x Therefore, the required answer is − 1 2xx Question #17 How to graph the derivative of a function when you are given the graph of the function? Answer: This may be accomplished by observing how the slope of the lines tangent to the graph alters when the value of x (the independent variable) varies. This is effective because the derivative provides a formula (a function) for calculating the tangent line slopes for different x values. The tangents on the left show a significant positive slope as we gaze from left to right. As we move past x=0 (the y-axis) and get closer to the high point at x=0.5, the slope declines and the tangents become more nearly level. Here, the y value is somewhat greater than 2, and it is referred to as a local or relative maximum value. As we move farther to the right, the tangent lines' slopes turn negative and drop until around x=1.5, at which point they flatten out once more (closer to horizontal). When we get to about 2.3, we hit a local minimum value and move on to a region of the graph where the tangent lines have a positive slope. Keep in mind that the function's local extreme values happen at the same x levels that result in the derivative being 0. (At the graph's x-intercept, the function's value is 0. If you want to experiment with this using a graphing tool and other features: try: f ( x ) = x5 − x 4 − 3x3 + x 2 − x and f  ( x ) = 5x 4 − 4 x3 − 9 x 2 + 2 x − 1 Question #18 A point is moving along the curve y = x in such a way that its x coordinate id increasing at the rate of 2 units per minute. At what rate is its slope changing (a) when x=1? (b) when x=4? Answer: Let's call our time variable t , in minutes. We know that: dx = 2.min −1 dt Additionally, we're attempting to determine how quickly the slope changes over time: d  dy    dt  dx  First, let's evaluate dy d = dx dx dy : dx ( x ) = 21x Now, we need to evaluate d  dy   : dt  dx  d  1  1 d 1  = dt  2 x  2  dt x  1  d 1   dx  =    2  dx x   dt  1 1 = − 3 2  2x 2 =− 1 2x 3 2   2.min −1   ( ) .min −1 Finally, all we have to do is plug in x =1 − 1 2x 3 2 1 x =4− 2x 3 2 x = 1 and x = 4 to answer the two parts to the problem. 1 .min −1 = − .min −1 2 .min −1 = − 1 1 .min −1 = − .min −1 2 8 16 Question #19 Find all points where the tangent line is horizontal: x + xy + y = 1 2 Answer: When y'=0, we know the tangent line is horizontal. 2 x + (1 y + xy ) + 2 y  y = 0 = 2 x + y + xy + 2 yy = 0 2x + y + x ( 0) + 2 y ( 0) = 0 x 2 + x ( −2 x ) + ( −2 x ) = 1y = −2 x 2 x2 − 2x2 + 4x2 = 1 3x 2 = 1 cx= 1 3 2  1  1 1 1 , −2  and  − , 2  3 3  3  3 Answer is   Question #20 How to find the critical numbers for h ( x ) = sin 2 x + cos x to determine the maximum and minimum? Answer: To begin, differentiate. h ( x ) = ( sin x ) + cos x 2 You can use the chain rule on ( sin x ) 2 h ( x ) = 2sin x cos x − sin x Critical numbers occur whenever the derivative equals 0 . Hence, 0 = 2sin x cos x − sin x 0 = sin x ( 2cos x − 1) If we solve, we get sin x = 0 or cos x = 1 2 This means that x = 0,  ,  5 3 , 3 Selecting test points in between will help us identify the points where the function is increasing or decreasing. Test point 1: x =  6       h   = sin   − sin   6 3 6 3 1   h   = − 6 2 2 3 −1   h   = 2 6   This is positive, so the function is increasing on  0, Test point 2: x =  2 If we evaluate within the derivative, we get:  . 3         h   = sin  2    − sin   2 2   2      h   = sin ( ) − sin   2 2   h   = 0 − 1 2   h   = −1 2 Hence,   h ( x ) is decreasing on  ,   3  Test point 3: x = 3 2  3   3   3  h   = sin  2  − sin    2   2   2   3 h   2   3   = sin ( 3 ) − sin     2   3 h   2   = 0 − ( −1)   3 h   2   =1  Hence,  5  h ( x ) is increasing on   ,   3  Accordingly, we can deduce that minimums will occur whenever and maximums will occur whenever x=  3  2 n AND x = 5  2 n . This is because a maximum is seen at the point where a function stops 3 increasing and begins decreasing, and a minimum is seen at a point where a function stops decreasing and begins increasing. A graphical verification yields the same result. 𝑔𝑟𝑎𝑝ℎ𝑦 = (𝑠𝑖𝑛𝑥)(𝑠𝑖𝑛𝑥) + 𝑐𝑜𝑠𝑥[−22.8,22.83, −11.4,11.38]