What is the factoring?

Find dy/dx by implicit differentiation. y cos x = 5x2 + 3y2

lbvjy [14]

Step 1:
Start by putting $\frac{d}{dx}$ in front of each term

$\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]$
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Step 2:

Deal with the terms in 'x' and the constant terms
$\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]$
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Step 3:

Use the chain rule for the terms in 'y'
$\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}$
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Step 4:

Use the product rule on the term in 'x' and 'y'
$(y) \frac{d}{dx} cos x+(cos x) \frac{d}{dx}y =10x+6y \frac{dy}{dx}
$
$y(-siny)+(cosx) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
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Step 5:

Rearrange to make $\frac{dy}{dx}$ the subject
$-y sin(y)+cos(x) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
$cos(x) \frac{dy}{dx}-6y \frac{dy}{dx}=10x+y sin(y)$
$[cos(x) - 6y] \frac{dy}{dx}=10x + y sin(y)$
$\frac{dy}{dx}= \frac{10x+ysin(y)}{cos(x)-6y}$ ⇒ Final Answer

5 0

3 years ago

Help me please i need your helpp

Inga [223]

Answer:

D) -2

Step-by-step explanation:

to identify the slope of a line written in slope-intercept form, it would be

the coefficient of the 'x' term

5 0

2 years ago

How many hundreths are equivalent to three-quaters<br><br>

8_murik_8 [283]

Answer:

7.5 hundredths (7 hundredths and 5 tenths) equals 3/4

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

What is the product of three square root of eight times four square root of three? Simplify your answer.

igor_vitrenko [27]

$The\ product\ of 3 \sqrt{8}*4 \sqrt{3}\ is\ 24 \sqrt{6}.$

First, recognize that all the numbers can be moved around because they are being multiplied together.
Next, rearrange the expression so that the whole-number coefficients are together:

$3\sqrt{8}*4\sqrt{6}$
$3*4*\sqrt{8}*\sqrt{3}$

Now try to simplify the square-roots to find any squares inside that can be reduced and taken out of the square-root:

$3*4*\sqrt{(2*2*2)}*\sqrt{3}$
$3*4*\sqrt{4*2}*\sqrt{3}$

Multiply the square roots together:
$3*4*\sqrt{(4*2*3)}$
$3*4*\sqrt{(4*6)}$

Now take out the square-root of 4 and simplify
$3*4*\sqrt{4}*\sqrt{6}$
$3*4*(2)*\sqrt{6}$
$24\sqrt{6}$

$Thus,\ your\ answer\ is\ 24\sqrt{6}$

7 0

3 years ago

Verify sine law by taking triangle in 4 quadrant<br>Explain with figure.<br>

Ksivusya [100]

Proof of the Law of Sines

The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

a

 sin  A

=

b

 sin  B

=

c

 sin  C

For more see Law of Sines.

Acute triangles

Draw the altitude h from the vertex A of the triangle

From the definition of the sine function

 sin  B =

h

c

a n d  sin  C =

h

b

or

h = c  sin  B a n d h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Repeat the above, this time with the altitude drawn from point B

Using a similar method it can be shown that in this case

c

 sin  C

=

a

 sin  A

Combining (4) and (5) :

a

 sin  A

=

b

 sin  B

=

c

 sin  C

- Q.E.D

Obtuse Triangles

The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

h

c

a n d  sin  C =

h

b

or

h = c  sin  B a n d h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Draw the second altitude h from B. This requires extending the side b:

The angles BAC and BAK are supplementary, so the sine of both are the same.

(see Supplementary angles trig identities)

Angle A is BAC, so

 sin  A =

h

c

or

h = c  sin  A

In the larger triangle CBK

 sin  C =

h

a

or

h = a  sin  C

From (6) and (7) since they are both equal to h

c  sin  A = a  sin  C

Dividing through by sinA then sinC:

a

 sin  A

=

c

 sin  C

Combining (4) and (9):

a

 sin  A

=

b

 sin  B

=

c

 sin  C

7 0

2 years ago