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galina1969 [7]
3 years ago
6

Find the derivative of ln(secx+tanx)

Mathematics
1 answer:
Sliva [168]3 years ago
3 0
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3000160

————————

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>

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Considering the slopes of the segments, the correct option is:

D. No, because the triangle ABC doesn't have a pair of perpendicular sides.

<h3>When are lines parallel, perpendicular or neither?</h3>

The slope, given by <u>change in y divided by change in x</u>, determines if the lines are parallel, perpendicular, or neither, as follows:

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Here, we have to find if there are perpendicular segments, that is, if two slopes multiplied have a value of -1, then:

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6 0
2 years ago
A high school has 44 players on the football team. The summary of the players' weights is given in the box plot. Approximately,
Marina CMI [18]

Answer:

Step-by-step explanation:

The missing image is attached below.

From the given information

The number of players is 44.

From the box plot;

The minimum weight = 154 pounds

The first quartile Q₁ = 159 which is 25% of the data below 159 pounds.

The second quartile Q₂ = 213 which is 50% of the data below 213 pounds

The third quartile Q₃ = 253 which is 75% of the data below 253 pounds

The maximum weight = 268 pounds.

Here, the value of 213 is the middle value and which signifies the median.

The 50% of the data are located both on the left side and the right side of the median.

Thus, the percentage of players weighting greater than or equal to 213 pounds is 50%.

7 0
3 years ago
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Answer:

answer is 11.9...... have a good day

6 0
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If you are a purist and want to solve the question the way it is written, you could do it this way.

dy/dx = d(1)/dx x^-2 - d(x^-2)/dx * 1
             ======================                    
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dy/dx =  - (-2 x^ - 3) / x^-4

dy/dx = 2 x^-3 * x^4
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