, meant to remind us of the definition of f (x) as the limit of difference quotients: = lim

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1 Mat 132 Differentiation Formulas Stewart 2.3 So far, we ave seen ow various real-world problems rate of cange and geometric problems tangent lines lead to derivatives. In tis section, we will see ow to solve suc problems by computing derivatives differentiating algebraically. Notations. We ave seen te Newton notation f x for te derivative of fx. Te alternative Leibnitz notation for te derivative is df dx, meant to remind us of te definition of f x as te it of difference quotients: f x df dx. Here fx+ fx, te difference in fx produced by te difference x+ x. Also, df and dx are meant to suggest very small and, but df dx is not literally te quotient of two small quantities, just a complicated symbol meaning te it of suc quotients. To illustrate: for fx x 2, te formula f x x 2 2x can be written in Leibnitz notation as: df dx d dx x2 2x. Te symbol df dx means te function f x; for a particular value of a derivative at x a, we write f a df. Te notation f Df is also used, and f x Dfx. xa dx Basic Derivatives. To compute derivatives witout a it analysis eac time, we use te same strategy as for its in Notes 1.6: we establis te derivatives of some basic functions, ten we sow ow to compute te derivatives of sums, products, and quotients of known functions. Teorem: i For a constant function fx c, we ave d dx c c 0. ii For fx x, we ave d dx x x 1. iii For fx x p wit p any real number, we ave: d dx xp x p px p 1. Proof: i and ii follow easily from te definition of f x. We prove iii in stages, for more and more general powers p, relying repeatedly on te family of formulas: a n b n a ba n 1 + a n 2 b + a n 3 b b n 1, valid for n 1, 2, 3,... First, we consider a wole number p n, and take a x+ and b x: x n x+ n x n x+ x x+ n 1 +x+ n 2 x+ +x n 1 x+ n 1 +x+ n 2 x+ +x n 1 x+0 n 1 +x+0 n 2 x+ +x n 1 nx n 1. Tus, x n nx n 1, and iii olds for p n. Second, we do a similar calculation for a negative integer p n, so tat x p 1 x ; in n te derivative it, we combine fractions and apply te a n b n formula wit a x and b x+. Te result simplifies to nxn 1 nx n 1. x 2n Notes by Peter Magyar magyar@mat.msu.edu is capital letter delta, te Greek D, standing for difference. Te small letter delta is δ.

2 Tird, we consider a fraction p n m wit m a wole number and n an integer, so tat x p x m n m x n. We take te derivative it wit numerator m x+ n m x n a b. As in 2.5 for 3 x, multiplying top and bottom by a m 1 + a m 2 b + a m 3 b b m 1 gets rid of te radicals m, leaving te numerator a m b m x+ n x n, wic we andled previously. Again, te it eventually simplifies to formula iii. Formula iii is also valid for an irrational power like p 2, but tis requires more teory: we will ave to wait until Calculus II to even state a clear definiton of x 2. Having computed all tese its, we never ave to do so again. Just from quoting te Teorem, we get formulas like: x 2 2x 1 2x; x 10 10x 9 ; 3 x x 1/3 1 3 x 2/ ; 1 x 2 x x 1 1x x 2 Derivative Rules. Suppose te functions fx, gx are differentiable at x, so tat f x and g x exist. Ten we get te following derivatives: Sum: fx + gx f x + g x. Difference: fx gx f x g x. Constant Multiple: c fx c f x for any constant c. Product: fxgx f xgx + fxg x. fx Quotient: f xgx fxg x gx gx 2, were gx 0. Te first tree of tese Rules, wic express te linearity of te derivative operation, are intuitive and easy to prove. For example te Sum Rule: fx + gx fx++gx+ fx+gx fx+ fx + gx+ gx fx+ fx gx+ gx + f x + g x. Here te tird equality follows from te Sum Law for its in Notes 1.6. Warning: Te derivative of a product is NOT te product of derivatives. We obtain te correct Product Rule from a geometric model: consider a rectangle wit canging sides of lengts fx and gx depending on some variable x, te upper left rectangle below: Te product fxgx is te area, and te derivative fxgx is te rate of cange of area wit respect to a cange in x. Suppose small increment produces some positive

3 increments fx+ fx and g gx+ gx in te sides, so tat te increment of area, g fx+gx+ fxgx, is te area of te tree edge rectangles: g gx + fx g + g. To get te derivative, we divide by to get te difference quotient, and send 0: fxgx g gx + fx gx + fx g g + + g g f xgx + fxg x + f x0 f xgx + fxg x. Note tat te vanising tird term corresponds to te tiny bottom rigt rectangle. Lastly, we prove te Quotient Rule: fx fx+ gx gx+ fx gx fx+gx fxgx+ gx+ gx fx+gx fxgx + fxgx fxgx+ gx+ gx Here, after putting te expression over a common denominator, we ave added and subtracted te quantity fxgx in te numerator, leaving te it uncanged. Our aim is to factor te first pair and last pair of terms: fx gx fx+ fx gx + fx gx gx+ gx+ gx fx+ fx 1 gx+ gx gx fx gx+ gx 1 f xgx fxg x f xgx fxg x gx+0 gx gx 2. We ave again used several Limit Laws from Notes 1.6. We could give anoter proof of te Product Rule in a very similar way. Derivative computations. By repeatedly using tese Rules, we can quickly compute te derivatives of most functions. example: Find x d dx x. Solution: x x 1/2 1 2 x1/ x 1/2 1 2 x, were we used te Basic Derivative x p px p 1 wit p 1 2. example: 10 0 since te derivative of any constant, even a complicated one, is zero. example: For fx 5x x 3, find te derivative f x df 5x 2 +1 x 3 5x 2 +1 x 3 + 5x 2 +1 x 3 by Product Rule 5x 2 +1 x 3 + 5x 2 +1 x 3 by Sum & Const Mult Rules 52x 1 +0 x 3 + 5x x 1/2 0 by Basic Derivatives 10x x 3 + 5x x dx : tidying up We can ceck tis formula algebraically for any fx, fx+, gx, gx+: just substitute for, g.

4 Note ow we used te derivative from te previous example, x 1 2 x 1/2. Anoter way to find te same derivative would be to multiply out first: fx 5x 2 +1 x 3 5x 2 x 15x 2 + x 3 5x 5/2 15x 2 + x 1/2 3. Ten we get te derivative: f x x5/ x x1/ x x 30x x. Tis agrees wit our previous answer, multiplied out. example: Differentiate gt t5 +1 t. Solution by te Quotient Rule: t g t dg dt t 5 +1 t t t5 +1 t t t 5 +1t t t t 2 5t4 t t t t1/2 t 3, were we use t t t 3/2. Solution by multiplying out: 1 t t t 3/2, so: gt t 5 +1t 3/2 t 7/2 + t 3/2 and g t 7 2 t5/2 3 2 t1/2. Example: A block of ice as lengt 10cm, widt 5cm, and eigt 20cm. Its lengt and widt are melting at a rate of 1cm per our, but its eigt is melting at 2cm per our because te ground is warmer tan te air. How fast is te volume decreasing? Solution: Te volume is V lw cm 3, were V, l, w, are all functions of time t. To get te rate of cange, we compute te derivative using te Product Rule twice, considering lw lw: dv dt V lw l w+lw l w+lw +w l w+lw +lw. We want te melt rate at te current time t 0, and we are given: l0 10 cm wit l 0 1 cm/r; w0 5 cm wit w 0 1 cm/r; and 0 20 cm wit 0 2 cm/r. Tus: V 0 l 0w00 + l0w 00 + l0w cm 3 /r. Higer derivatives. Since te derivative operation turns a function fx into anoter function f x, we can do it again to f x, obtaining yet anoter function denoted f x f x or d2 f dx d df dx dx, called te second derivative of fx. In real-world terms, if f x is te rate of cange of fx, ten f x is te rate of cange of f x, namely ow muc te rate f x is speeding up or slowing down. Example: A stone falls ft 16t 2 ft in t seconds. Compute te repeated derivatives of tis function, and interpret teir pysical meaning. Te first derivative is f t 16t 2 162t 1 32t ft/sec. vt f t 32t ft/sec, increasing proportional to time. Tis is te velocity Te second derivative is f t 32t 32, wit units ft/sec per sec ft/sec 2. It means te rate of cange of velocity, ow many ft/sec of speed is gained eac second. Tis is te acceleration of te stone, at f t 32 ft/sec 2, te constant acceleration due to gravity.

5 Te tird derivative f t 32 0, meaning te rate of cange of a constant acceleration is zero. Te pysics term for tis quantity is te jerk, and since te jerk ere is zero, we see tat gravity does not jerk: it pulls smootly. All iger derivatives are also zero; tese do not ave names.