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tutorial 1 solution

# tutorial 1 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA 1505 Mathematics I Tutorial 1 (Solution Notes) 1. Let f ( x )= 6 x and g ( x ± | 3 x | . Find an expression for ( g f )( x ) ( f g )( x ). Solution By the de±nition of function composition , we know that ( g f )( x g [ f ( x )] , ( f g )( x f [ g ( x )] , then ( g f )( x g [ 6 x ]= ² | 3 6 x | , ( f g )( x f [ ± | 3 x | 6 ± | 3 x | . Hence, ( g f )( x ) ( f g )( x ² | 3 6 x |− 6 ± | 3 x | . That is the answer. ± Tips If you are not familiar with function composition, check this web page: http://en.wikipedia.org/wiki/Composite function. 1

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2. Find the frst derivatives o± the ±ollowing ±unctions. (a) y = ax + b cx + d (b) y =sin n x cos( mx ) (c) y = e x 2 + x 3 (d) y = x 3 4( x 2 + e 2 +ln2) (e) y = ± sin θ cos θ 1 ² 2 (±) y = t tan(2 t )+7 (g) r =sin( θ + θ +1) (h) s = 4 cos x + 1 tan x Preliminaries This question is testing your some basic rules o± frst derivatives: First o± all, please notice that there are three kinds o± notations ±or di²erentiation that are widely used nowadays, and will appear in the lectures, tutorials, and tests o± MA1505 and MA1506 ±requently: Leibniz’s Notation: dy dx . Lagrange’s Notation: y ± ( x ) . Newton’s Notation: ˙ y ( x ) . It is no use to distinguish these notations, just keeping in mind that all o± the three notations are the same. For more in±ormation, please re±er to: http://en.wikipedia.org/wiki/Derivative. (1) Quotient Rule I± the ±unction is o± the ±orm f ( x )= A ( x ) B ( x ) ,then f ± ( x [ A ( x )] ± B ( x ) A ( x ) [ B ( x )] ± B ( x ) B ( x ) . Abbreviation: ± A B ² ± = A ± B AB ± B 2 . (2) Product Rule I± the ±unction is o± the ±orm f ( x A ( x ) B ( x ), then f ± ( x )=[ A ( x )] ± B ( x )+ A ( x ) [ B ( x )] ± . Abbreviation: ( AB ) ± = A ± B + AB ± . 2
(3) Chain Rule There are two ways to express this rule, they are equivalent, and both of them are very useful. You can choose your favorite. (1) ( f g ) ± ( x )= f ± ( g ( x )) g ± ( x ) ; (2) dy dx = dy du · du dx . (4) Constant Rule If the function is a constant function, then its derivative is 0. f = Constant = f ± =0 . You can check your lecture notes for more details. All of them are both very important rule, and you need to be familiar with them. Solution (a) Qn: y = ax + b cx + d , Ans: Use Quotient Rule: ± A B ² ± = A ± B AB ± B 2 . Then, y ± = ( ax + b ) ± ( cx + d ) ( ax + b )( cx + d ) ± ( cx + d ) 2 = a ( cx + d ) c ( ax + b ) ( cx + d ) 2 = ad bc ( cx + d ) 2 . Here, the Frst equality is using quotient rule, the second is using constant rule, and the last gets by some basic algebraic operations on the numerator. (b) Qn: y =sin n x cos mx , Ans: Use Product Rule: ( AB ) ± = A ± B + AB ± , we get: y ± =[sin n x · cos mx ] ± n x ] ± · cos mx +sin n x · [cos mx ] ± Then use Chain Rule: ( f g ) ± ( x f ± ( g ( x )) g ± ( x ) , we know: [sin n x ] ± = n sin n 1 x, [cos mx ] ± = m sin mx. Hence, y ± = n sin n 1 x cos x cos mx m sin n x sin mx. 3

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(c) Qn: y = e x 2 + x 3 Ans: y ± = e x 2 + x 3 (2 x +3 x 2 ) (use chain rule) (d) Qn: y = x 3 4( x 2 + e 2 +ln2), Ans: y ± =3 x 2 8 x, (note that e 2 and ln 2 are constants) (e) Qn: y = ± sin θ cos θ 1 ² 2 Ans: 2sin θ (cos θ 1) 2 (use quotient and chain rule) (f) Qn: y = t tan(2 t )+7 Ans: t sec 2 (2 t )+tan(2
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tutorial 1 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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