Directions: For each category see if you can devise the rule(s) that was used to

Question

Directions: For each category see if you can devise the rule(s) that was used to find the derivative(s) and then use it to find the derivatives of the functions that follow.

Category 1: Sine and Cosine

Function: f(x)=sin?(2x) 3. Function: f(x)=3sin?(x^2)
Derivative: f^' (x)=2cos?(2x) Derivative: f^' (x)=6xcos(x^2)




Function: f(x)=cos?(5x) 4. Function: f(x)=cos?(x^3)
Derivative: f^' (x)=-5sin?(5x) Derivative: f^' (x)=-3x^2 sin?(x^3)


What is the formula for the derivative of Sine? What about for Cosine? What role does the chain rule play in these formulas?











Find the derivatives of the following:

f(x)=2sin?(3x-2) 2. f(x)=-cos?(x^2-5x)















Category 2: Tangent and Cotangent

Function: f(x)=tan?(3x) 3. Function: f(x)=7tan?(x^2)
Derivative: f^' (x)=3?sec?^2 (3x) Derivative: f^' (x)=14x?sec?^2 (x^2)



Function: f(x)=cot?(6x) 4. Function: f(x)=1/3 cot?(2x^3)
Derivative: f^' (x)=-6?csc?^2 (6x) Derivative: f^' (x)=-2x^2 ?csc?^2 (2x^3)



What is the formula for the derivative of Tangent? What about for Cotangent?







Find the derivatives of the following:

f(x)=1/2 tan?(x^4) 2. f(x)=6cot?(x^3-2x^2)
























Category 3: Secant and Cosecant

1. Function: f(x)=sec?(8x) 3. Function: f(x)=csc?(9x)
Derivative: f^' (x)=8 sec?(8x) tan?(8x) Derivative: f^' (x)=-9 csc?(9x) cot?(9x)


2. Function: f(x)=2/5 sec?(3x^5) 4. Function: f(x)=2csc?(x^6)
Derivative: f^' (x)=6x^4 sec?(3x^5 )tan?(3x^5) Derivative: f^' (x)=-12x^5 csc?(x^6 )cot?(x^5)


What is the formula for the derivative of Secant? What about for Cosecant?







Find the derivatives of the following:

f(x)=-2sec?(x^2) 2. f(x)=4csc?(2x^3-1)


























Combining Rules

If f(x)=xsin(3x) what rule that we have previously learned should we use to find its derivative? Use that rule to find the derivative of f(x).






If f(x)=?tan?^3 (5x) what rule that we have previously learned should we use to find its derivative? Use that rule to find f

Explanation / Answer

Category 1: Sine and Cosine Function: f(x)=sin?(2x) 3. Function: f(x)=3sin?(x^2) Derivative: f^' (x)=2cos?(2x) Derivative: f^' (x)=6xcos(x^2) ans= The easiest thing to do would be to find the derivative of sin^3(x^2) first: sin^3(x^2)=[(sin(x^2)]^3 d/dx=3[sin(x^2)]^2*[2xcos(x^2)] d/dx=6xcos(x^2)sin^2(x^2) Now it's easier to derive the entire function: f(x) = [1+sin^3(x^2)]^5 f'(x)=5[1+sin^3(x^2)]^4*(6xcos(x^2)sin… 0 , f(x) is increasing on that interval f'(x)=5x^4-5 (-8,-1): choose x=-2; f'(-2)= 75 > 0 : f(x) is increasing on (-8,-1) (-1.1): choose x=.5; f'(.5)= -4.69 < 0 : f(x) is decreasing on (-1,1) (1, 8): choose x=2; f'(2)= 75 > 0 : f(x) is increasing on (1, 8) b) f'(x) = 5x^4-5 f''(x) = 20x^3 At x=-1 , f''(x) = -20 < 0, so f has a relative maximum At x=1, f''(x) = 20 > 0, so f has a relative minimum c) f''(x) =20x^3 =0 x=0 --- is a point of inflection Test for concavity: Consider the intervals (-8, 0),(0,8 ) choose any one point from each of the intervals. if f''(x) < 0 , f(x) is concave down on that interval if f''(x) > 0 , f(x) is concave up on that interval f''(x)=20x^3 (-8, 0): choose x=-2; f''(-2)=-160 < 0, f is concave down on (-8, 0) (0, 8): choose x=2; f''(2)= 160 > 0, f is concave up on (0, 8) d) f''(x) =20x^3 =0 x=0 --- is a point of inflection

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.