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3 years ago

Derivative using chain rule of cosx/1+sinx

1 answer:

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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The amount of nuts in each bag will be less than 7/8 because 7/8 is divided by 4.

Gina will be able to put 7/32 pounds of nuts in each bag.

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3 years ago

What is the circumstance of the circle

Alex_Xolod [135]

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Explanation:

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3 years ago

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diamong [38]

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Step-by-step explanation:

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3 years ago

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3 years ago

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

8 0

4 years ago