HW5solutions - HW 5 Solutions Math 1A with Prof. Stankova...

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Unformatted text preview: HW 5 Solutions Math 1A with Prof. Stankova UC-Berkeley, Spring 2009 3.3: Derivatives of Trigonometric Functions 2. We have f ( x ) = x sin x . We are asked to find the derivative f ( x ). We use the product rule and then the power rule and the fact that the derivative of sin x is cos x : f ( x ) = x sin x f ( x ) = ( x sin x ) = ( x ) sin x + x ( sin x ) = ( x 1 / 2 ) sin x + x ( sin x ) = 1 2 x- 1 / 2 sin x + x ( sin x ) = 1 2 x- 1 / 2 sin x + x cos x = sin x 2 x + x cos x = sin x + 2 x cos x 2 x 10. We are asked to find the derivative of y = 1 + sin x x + cos x . We use the quotient rule, and the formulas for the derivative of sine and cosine: sin x = cos x , cos x =- sin x . We then simplify using the Pythagorean identity 1 = cos 2 x +sin 2 x . It is a matter of taste whether to expand the denominator; for the readers convenience, we report both possibilities. Warning: one cannot cancel x s and cos x s between the numerator and the denominator, because of the +s in the denominator. y = 1 + sin x x + cos x y = 1 + sin x x + cos x = ( 1 + sin x ) ( x + cos x )- ( 1 + sin x )( x + cos x ) ( x + cos x ) 2 = ( 0 + cos x )( x + cos x )- ( 1 + sin x )( 1- sin x ) ( x + cos x ) 2 = x cos x + cos 2 x- (1- sin 2 x ) ( x + cos x ) 2 = x cos x + cos 2 x + sin 2 x- 1 ( x + cos x ) 2 = x cos x + 1- 1 ( x + cos x ) 2 = x cos x ( x + cos x ) 2 1 16. We are asked to find the derivative of y = x 2 sin x tan x . We use the product rule twice, and recall that sin x = cos x and tan x = 1 / cos 2 x : y = x 2 sin x tan x y = ( x 2 sin x tan x ) = ( x 2 sin x ) tan x + x 2 sin x tan x = ( x 2 ) sin x + x 2 sin x tan x + x 2 sin x tan x = ( x 2 ) sin x tan x + x 2 sin x tan x + x 2 sin x tan x = 2 x sin x tan x + x 2 cos x tan x + x 2 sin x 1 cos 2 x = 2 x sin x tan x + x 2 sin x + x 2 sin x cos 2 x 24. The exercises asks for the tangent line at (024....
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HW5solutions - HW 5 Solutions Math 1A with Prof. Stankova...

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