(x,y) 4. Calculus I: Differentiation

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1 4. Calculus I: Differentiation 4. The eriatie of a function Suppose we are gien a cure with a point lying on it. If the cure is smooth at then we can fin a unique tangent to the cure at : If the tangent is unique then the graient of the cure at is efine to be the graient of the tangent to the cure at. The process of fining the general graient function for a cure is calle ifferentiation. B a b c Consier the chor B. s B gets closer to, the graient of the chor gets closer to the graient of the tangent at. Here the cure in a) is smooth at, but the cures in b) c) are not. nton Co City Uniersity) M60 Week 4 utumn 007 / 37 nton Co City Uniersity) M60 Week 4 utumn 007 / 37 For y f ), the graient function is efine by ) δy ) f + ) f ) lim lim ,y+ δy),y) We enote the graient function by or f ), call it the eriatie of f. This is not the formal efinition of the eriatie, as we hae not eplaine eactly what we mean by the limit as 0. But this intuitie efinition will be sufficient for the basic functions which we consier. δy Eample 4..: Take f ) c, a constant function. t eery the graient is 0, so f ) 0 for all. Or f + ) f ) Eample 4..: Take f ) a. c c 0. t eery the graient is a, so f ) a for all. Or f + ) f ) a + ) a a a. nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 Eample 4..3: Take f ). Now we nee to consier the secon formulation, as we cannot simply rea the graient off from the graph. f + ) f ) + ) + + ) + ) +. The limit as tens to 0 is, so f ). Eample 4..4: Take f ). f + ) f ) + ) + ) ) + ) ) + ) + ). The limit as tens to 0 is, so f ). nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 Eample 4..5: Take f ) n with n N n >. Recall that so a n b n a b)a n + a n b + a n 3 b + + b n ) a n b n a b an + a n b + a n 3 b + + b n where the sum has n terms. s a b we hae a n b n ) lim lim a n + a n b + a n 3 b + + b n ) nb n. a b a b a b If a + b then ) f + ) f ) lim lim 0 a b Hence f ) n n. a n b n a b ) nb n n n. Eample 4..6: f ) sin. We use the ientity for sin + sin B. so f + ) f ) sin f + ) f ) sin ) ) cos + ) cos + ). We nee the following fact which we will not proe here): so sin θ lim θ 0 θ f sin ) ) lim cos + ) cos). 0 nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37

2 Some star eriaties, which must be memorise: f ) f ) k k k e e ln sin cos cos sin tan sec cosec cosec cot sec sec tan cot cosec Some of these results can be erie from the results in the following sections, or from first principles. Howeer it is much more efficient to know them. 4. Differentiation of compoun functions Once we know a few basic eriaties, we can etermine many others using the following rules: Let u) ) be functions of, a b be constants. Function Deriatie Sum ifference au ± b a u ± b Prouct u u + u u Quotient u u u Composite u)) z. z where z ). The final rule aboe is known as the chain rule has the following special case ua + b) a u a + b) For eample, the eriatie of sina + b) is a cosa + b). nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 Eample 4..: Differentiate Eample 4..: Differentiate y y +. + ) ) 4 + ) + ). Eample 4..3: Differentiate y ln + 3). ln + 3) Eample 4..4: Differentiate y e 5. Set z 5, then z z ez 5 5e 5. nton Co City Uniersity) M60 Week 4 utumn 007 / 37 nton Co City Uniersity) M60 Week 4 utumn 007 / 37 Eample 4..5: Differentiate y 4 sin + 3). Set z + 3, then z 4 cosz) 8 cos + 3). z s we hae alrea note, some of the star eriaties can be euce from the others. Eample 4..6: Differentiate y tan sin cos. cos cos sin sin ) cos cos + sin cos cos sec. Eample 4..7: y cosec sin. sin.0). cos sin cos sin cosec cot. Eample 4..8: y ln + + ), i.e. y ln u where u + +. u u so ) u + + ). ) nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 Eample 4..9: y. y e ln ) e ln ), i.e. y e u where u ln. u eu e ln ln) + ) ln) + ). 4.3 Higher eriaties The eriatie is itself a function, so we can consier its eriatie. If y f ) then we enote the secon eriatie, i.e. the eriatie of with respect to, by y or f ). We can also calculate the higher eriaties n y or f n) ). n Eample 4.3.: y ln + ). Let z. + y z + ). ) + ) ) + ). Eample 4.3.: Show that y e sin) satisfies y + + 5y 0. e sin + e cos e cos sin ) y e cos sin ) + e 4 sin cos ) e 3 sin 4 cos ). nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37

3 Writing s for sin c for cos we hae y + y + 5y e 3s 4c s + 4c + 5s) 0. Now 4 + ) ) Eample 4.3.3: Ealuate ) 3 + ) + 3) at 0. We coul use the quotient rule, but this will get complicate. Instea we use partial fractions. y ) + 3) + + B + ) + C + 3. y + ) ) 3 3 y ) ) 4 substituting 0 we obtain that 3 y 0) We obtain check!) 0, B, C 3. nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 Generally it is har to gie a simple formula for the nth eriatie of a function. Howeer, in some cases it is possible. The following can be proe by inuction. Eample 4.3.4: y e a. We can show that aea n y n an e a. y a e a. Eample 4.3.5: y sina). We can show that y a cosa) a sina + π ) y a sina) a sina + π) y a 3 cosa) a 3 sina + 3π ) y i) a 4 sina) a 4 sina + π). n y n an sina + nπ ). nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 Sometimes it is useful to use the Leibnitz rule, which gies a formula for the nth eriatie of a prouct of functions. n n fg) f )g + n n So for eample 3 ) n n f ) n g)+ ) n + n 3 fg) f )g f ) g) + 3 ) n n f ) n g)+ n n f ) g) + f n n g). 3 f ) g) + f 3 g). 4.4 Differentiating implicit functions Sometimes we cannot rearrange a function into the form y f ), or we may wish to consier the original form anyway for eample, because it is simpler). Howeer, we may still wish to ifferentiate with respect to. Gien a function gy) we hae from the chain rule gy)) gy)). s can be seen, this is ery similar to the binomial theorem, can also be proe by inuction using the prouct rule). nton Co City Uniersity) M60 Week 4 utumn 007 / 37 nton Co City Uniersity) M60 Week 4 utumn 007 / 37 Eample 4.4.: + 3y y 4. Therefore we hae + 3y y 4 ) ) y ) y 4 ) 0 + 3y + 3 y ) 4y 3 + 3y + 6y 3 4y 0 0. Eample 4.4.: + 3 y. Therefore we hae + 3 y ) ) ) 3 y y nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37

4 4.5 Differentiating parametric equations Sometimes there is no easy way to epress the relationship between y irectly in a single equation. In such cases it may be possible to epress the relationship between them by writing each in terms of a thir ariable. We call such equations parametric equations as both y epen on a common parameter. Eample 4.5.: t 3 y t 4t +. lthough we can write this in the form y the parametric ersion is easier to work with. To ifferentiate a parametric equation in the ariable t we use Eample 4.5.: Continue.) so t 4 t 4 3t.. 3t nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 Eample 4.5.: Fin the secon eriatie with respect to of Therefore Now y sin θ y cos θ. θ cos θ sin θ. θ sin θ 4 sin θ. cos θ ) 4 sin θ) θ 4 cos θ 4 sin θ) 4. θ cos θ Note: The rules so far may suggest that eriaties can be treate just like fractions. Howeer y y t in general. Moreoer y ). nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / Tangents normals to cures alrea efine the alue of the eriatie f of a function f at a point 0 to be the graient of f at 0. Thus we can easily use the eriatie to write own the equation of the tangent to that point. Using the equation for a line passing through 0, f 0 )) we hae that the tangent to f at 0 is y f 0 ) 0) 0 ). The normal to f at 0 is the line passing through 0, f 0 )) perpenicular to the tangent. This has equation when this makes sense). y f 0 ) 0) 0) Eample 4.6.: Fin the equation of the tangent normal to the cure y at the point, 3). 6 hence ) 4 6. Hence the equation of the tangent is y + 3 ) i.e. y +. The graient of the normal is, hence the equation of the normal is y + 3 ) i.e. y 4. nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / Stationary points points of infleion We can tell a lot about a function from its eriaties. Eample 4.7.: If f ) > 0 for a < < b then f is increasing on a < < b e.g. arcs PQ, SU, UV. P Q R T U V If f ) < 0 for a < < b then f is ecreasing on a < < b e.g. arc QS. S nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37

5 stationary point on a cure y f ) is a point 0, f 0 )) such that f 0 ) 0. These come in arious forms: Type f ) Test f ) Local maimum Changes from + to e Local minimum Changes from to + +e Point of infleion No sign change see below) e.g. Q is a ma, S is a min, U is a point of infleion. point of infleion is one where f 0 ) 0 f changes sign at 0. e.g. R, T, U. If f ) > 0 for a < < b then f is concae up on a < < b e.g. arc RST. If f ) < 0 for a < < b then f is concae own on a < < b e.g. arc PQR. nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 Note that the maima minima aboe are only local. This means that in a small region about the gien point they are etremal alues, but perhaps not oer the whole cure. Etremal alues for the whole cure are calle global maima or minima. Eample 4.7.: Consier the function f on the omain X Y gien by the graph X B C D Y Eample 4.7.3: Fin the stationary alues points of infleion of y y Stationary points when 0, i.e. check),,. Both C are local maima, B D are local minima. Howeer the global maimum is at Y the global minimum at X. nton Co City Uniersity) M60 Week 4 utumn / 37 nton Co City Uniersity) M60 Week 4 utumn / 37 y y, 7) Min, 5) Ma, 0) Min Points of infleion at 3 ± 7), i.e., y) 0., 3.36), y).55,.3). nton Co City Uniersity) M60 Week 4 utumn / 37