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DIA [1.3K]
3 years ago
13

(a) Use the Quotient Rule to differentiate the function f(x)=tan(x)-1/sec(x). f'(x)=

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
4 0

Answer:

(a) sin(x) + cos(x)

(b) sin(x) + cos(x)

(c) Both answers are equivalent

Step-by-step explanation:

(a) The given function is:

f(x) = \frac{tan(x)-1}{sec(x)}

According to the quotient rule:

f(x) = \frac{g(x)}{h(x)}\\f'(x)= \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}

Applying the quotient rule:

f(x) = \frac{tan(x)-1}{sec(x)}\\f'(x)=\frac{sec^2(x)*sec(x)-(tan(x)-1)*sec(x)tan(x)}{sec(x)^2}\\f'(x)=\frac{sec^3(x)-sec(x)tan^2(x)+sec(x)tan(x)}{sec(x)^2}\\f'(x)=sec(x)+\frac{tan(x)-tan^2(x)}{sec(x)}\\ \frac{tan(x)}{sec(x)}=sin(x) \\f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\

This can be simplified to:

f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\f'(x) = \frac{1}{cos(x)}+sin(x)-\frac{sin^2(x)}{cos(x)}\\f'(x)=\frac{1+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{sin^2(x)+cos^2(x)+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{cos^2(x)+sin(x)cos(x)}{cos(x)}\\ f'(x)=sin(x) +cos(x)

(b) Simplifying in terms of sin(x) and cos(x):

f(x) = \frac{tan(x)-1}{sec(x)}\\f(x)=\frac{\frac{sin(x)}{cos(x)}-1 }{\frac{1}{cos(x)} } \\f(x)=sin(x)-cos(x)\\f'(x) = cos(x)+sin(x)

(c) As proven above, both answers are equivalent.

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Question # 13

Answer:

The required equation for the given function is <em>y = 4sin(x/2+2π/3) -2 , as shown attached graph diagram.</em>

<em>Step-by-step explanation: </em>

As the general sine function is given by

y=asin(bx+c)+d.......[A]

  • amplitude = a
  • period = 2π ÷ b
  • Phase shift = -c ÷ b
  • Vertical shift = d

As in the question,  

  • amplitude = a = 4
  • period = 4π
  • phase shift = -4π/3
  • Vertical shift = d = -2

As  

period = 2π ÷ b  

b = 2π/period

b = 2π/4π ∵ period = 4π

b = 1/2  

Also

Phase shift = -c/b

-4π/3 = -c/b ∵ phase shift = -4π/3

4π/3 = c/b  

c = b × 4π/3  

c = 1/2 × 4π/3  

c = 4π/6  

c = 2π/3

So, putting Amplitude ⇒ a = 4, Vertical shift ⇒ d = -2, b = 1/2 ,  

and c = 2π/3 in Equation [A] would bring us the required equation for the given function.

y=asin(bx+c)+d

y = 4sin(x/2+2π/3)+(-2)

y = 4sin(x/2+2π/3) -2            

<em>Note: The graph is also shown in attached diagram.</em>

                                             Question # 14

<em>Answer:</em>

The required equation for the given function is y = cot(x+π/3)+2, as shown in attached graph diagram.

<em>Step-by-step explanation: </em>

As the general cotangent function is given by

y=acot(bx+c)+d.......[A]

  • amplitude = a
  • period = π ÷ b
  • Phase shift = -c ÷ b
  • Vertical shift = d

As in the question,  

  • period = π
  • phase shift = -π/3
  • Vertical shift = d = 2

As  

period = π ÷ b  

b = π/period

b = π/π ∵ period = 4π

b = 1  

Also

Phase shift = -c/b

-π/3 = -c/b ∵ phase shift = -π/3

π/3 = c/b  

c = b × π/3  

c = 1 × π/3  

c = π/3

So, putting vertical shift ⇒ d = 2, b = 1 and   c = π/3 in Equation [A] would bring us the required equation for the given function.

y=acot(bx+c)+d

y = cot(x+π/3)+2

<em>Note: The graph is also shown in attached diagram.</em>

Keywords: amplitude, period , phase shift , vertical shift

Learn more about trigonometric functions of equations from brainly.com/question/2643311

#learnwithBrainly

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