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3 years ago

I NEED HELP PLEASE, THANKS! :)

Mathematics

1 answer:

Elena L [17]3 years ago

7 0

Answer:

First option is the right choice.

Step-by-step explanation:

$\frac{1}{\cos \left(x\right)+1}+\frac{1}{\cos \left(x\right)-1}\\\\\frac{\cos \left(x\right)-1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}+\frac{\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{\cos \left(x\right)-1+\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{2cos\left(x\right)}{cos^2\left(x\right)-1}\\\\=\frac{2cos\left(x\right)}{sin^2\left(x\right)}$

$=2\cdot \frac{cos\left(x\right)}{sin\left(x\right)}\cdot \frac{1}{sin\left(x\right)}\\\\=-2cot(x)csc(x)$

Best Regards!

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Find the derivative of ln(secx+tanx)

Sliva [168]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000160

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Find the derivative of

$\mathsf{y=\ell n(sec\,x+tan\,x)}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1}{cos\,x}+\dfrac{sin\,x}{cos\,x} \right )}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1+sin\,x}{cos\,x} \right )}$

You can treat y as a composite function of x:

$\left\{\! \begin{array}{l} \mathsf{y=\ell n\,u}\\\\ \mathsf{u=\dfrac{1+sin\,x}{cos\,x}} \end{array} \right.$

so use the chain rule to differentiate y:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(\ell n\,u)\cdot \dfrac{d}{dx}\!\left(\dfrac{1+sin\,x}{cos\,x}\right)}$

The first derivative is 1/u, and the second one can be evaluated by applying the quotient rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{\frac{d}{dx}(1+sin\,x)\cdot cos\,x-(1+sin\,x)\cdot \frac{d}{dx}(cos\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(0+cos\,x)\cdot cos\,x-(1+sin\,x)\cdot (-\,sin\,x)}{(cos\,x)^2}}$

Multiply out those terms in parentheses:

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos\,x\cdot cos\,x+(sin\,x+sin\,x\cdot sin\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos^2\,x+sin\,x+sin^2\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(cos^2\,x+sin^2\,x)+sin\,x}{(cos\,x)^2}\qquad\quad (but~~cos^2\,x+sin^2\,x=1)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}$

Substitute back for $\mathsf{u=\dfrac{1+sin\,x}{cos\,x}:}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{~\frac{1+sin\,x}{cos\,x}~}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{cos\,x}{1+sin\,x}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}$

Simplifying that product, you get

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+sin\,x}\cdot \dfrac{1+sin\,x}{cos\,x}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{cos\,x}}$

∴ $\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=sec\,x} \end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>derivative composite function logarithmic logarithm log trigonometric trig secant tangent sec tan chain rule quotient rule differential integral calculus</em>

3 0

3 years ago

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andre [41]

Option C would be the correct answer :)

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2 years ago

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Hello here is a solution ;

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2 years ago

Is there enough information to conclude that the two triangles are congruent? If so, what is a correct congruence statement?

OlgaM077 [116]

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5 0

3 years ago

Read 2 more answers

Amir stands on a balcony and throws a ball to his dog who is at ground level. The ball's height, in meters above the ground, aft

telo118 [61]

Answer:

Time t = 2 seconds

It will reach the maximum height after 2 seconds

Completed question;

Amir stands on a balcony and throws a ball to his dog who is at ground level. The ball's height, in meters above the ground, after t seconds that Amir has thrown the ball is given by:

H (t) = -(t-2)^2+9

many seconds after being thrown will the ball reach its maximum height?

Step-by-step explanation:

The equation of the height!

h(t) = -(t-2)^2 + 9 = -(t^2 -4t +4) + 9

h(t) = -t^2 +4t -4+9

h(t) = -t^2 + 4t +5

The maximum height is at dh/dt = 0

dh/dt = -2t +4 = 0

2t = 4

t = 4/2 = 2

Time t = 2 seconds

It will reach the maximum height after 2 seconds

8 0

3 years ago