234_James Stewart Calculus 5 Edition Answers

234_James Stewart Calculus 5 Edition Answers - h f ( x + h...

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dy dx = sec x sec 2 x ± (tan x ± 1)sec x tan x sec 2 x = sec x sec 2 x ± tan 2 x +tan x ( ) sec 2 x = 1+tan x sec x y y =sin x ± cos x ² y / =cos x +sin x y = sin x x 2 ² y / = x 2 cos x ± (sin x )(2 x ) x 2 ( ) 2 = x ( x cos x ± 2sin x ) x 4 = x cos x ± 2sin x x 3 y = ± ( ± +cot ± ) ² y / = ± (1 ± csc 2 ± )+( ± +cot ± )( ± ± cot ± )= ± (1 ± csc 2 ± ± ± cot ± ± cot 2 ± ) = ± ( ± cot 2 ± ± ± cot ± ± cot 2 ± ) {1+cot 2 ± =csc 2 ± } = ± ( ± ± cot ± ± 2cot 2 ± )= ± ± cot ± ( ± +2cot ± ) y =sec ± tan ± ² y / =sec ± (sec 2 ± )+tan ± (sec ± tan ± )=sec ± (sec 2 ± +tan 2 ± ) 1+tan 2 ± =sec 2 ± sec ± (1+2tan 2 ± ) sec ± (2sec 2 ± ± 1) y = fgh y / = f / gh + fg / h + fgh / y = x sin x cos x ² dy dx =sin x cos x + x cos x cos x + x sin x ( ± sin x )=sin x cos x + x cos 2 x ± x sin 2 x d dx csc( x ) ( ) = d dx 1 sin x = (sin x )(0) ± 1(cos x ) sin 2 x = ± cos x sin 2 x = ± 1 sin x ³ cos x sin x = ± x cot x d dx sec x ( ) = d dx 1 cos x = (cos x )(0) ± 1( ± sin x ) cos 2 x = sin x cos 2 x = 1 cos x ³ sin x cos x =sec x tan x d dx cot x ( ) = d dx cos x sin x = (sin x )( ± sin x ) ± (cos x )(cos x ) sin 2 x = ± sin 2 x +cos 2 x sin 2 x = ± 1 sin 2 x = ± 2 x f ( x )=cos x ² f / ( x ) = lim
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Unformatted text preview: h f ( x + h ) f ( x ) h = lim h cos ( x + h ) cos x h = lim h cos x cos h sin x sin h cos x h Another method: Simplify first: . 13. 14. 15. Using the identity , we can write alternative forms of the answer as or 16. Recall that if , then . 17. 18. 19. 20. 2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions...
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