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almond37 [142]
3 years ago
14

Derivative using chain rule of cosx/1+sinx

Mathematics
1 answer:
sleet_krkn [62]3 years ago
4 0
I am not sure how you would approach this using the chain rule, but I can use the quotient rule. 

The derivative of cos(u)=-sin(u)
The derivative of sin(u)=cos(u)

The quotient rule states:

[f'(x)g(x)-f(x)g'(x)] / [g(x)]^2

f'(x)= -sin(x)
g(x)= [1+sin(x)]
f(x)= cos(x)
g'(x)= cos(x)

Plug in:

[-sin(x)*[1+sin(x)] - [cos(x)*cos(x)] / (1+sin(x))^2

Simplify by distributing 

-sin(x) - sin^2(x) - [cos(x)*cos(x)] / (1+sin(x))^2

Multiply the cos(x)*cos(x)

-sin(x) - sin^2(x) - cos^2(x) / (1+sin(x))^2

That is the answer or the answer could be simplified further by pulling out the negative.

- [sin(x)+sin^2(x)+cos^2(x)] / (1+sin(x))^2
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swat32

Answer: (10,-16)

Step-by-step explanation:

A(-4,8)\ \ \ \ C(3,-4)\ \ \ \ \ B(x_B,y_B)=?\\

See the graph below.

                             \displaystyle\\\boxed {x_C=\frac{x_A+x_B}{2} }\\\\

Multiply both parts of the equation by 2:

2x_C=x_A+x_B\\2x_C-x_A=x_A+x_B-x_A\\2x_C-x_A=x_B

Hence,

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                            \displaystyle\\\boxed {y_C=\frac{y_A+y_B}{2} }

Multiply both parts of the equation by 2:

2y_C=y_A+y_B\\2y_C-y_A=y_A+y_B-y_A\\2y_C-y_A=y_B

Hence,

y_B=2*(-4)-8\\y_B=-8-8\\y_B=-16\\Thus,\ B(10,-16)

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The time period of a simple pendulum is directly related to the square root of its length. An 8 foot pendulum has a period of 3.
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T_1=k \sqrt{8}

T_2=k \sqrt{3}

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Then:

T_2=3.2 \sqrt{ \frac{3}{8} } =1.96 secs


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