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bezimeni [28]
3 years ago
7

What is the value of the variable is

Mathematics
1 answer:
Gala2k [10]3 years ago
6 0
The value of the variable is what the variable equals to.
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Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,
Anton [14]

Answer:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}

Step-by-step explanation:

So we have the equation:

\tan(x-y)=\frac{y}{8+x^2}

And we want to find dy/dx.

So, let's take the derivative of both sides:

\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]

Let's do each side individually.

Left Side:

We have:

\frac{d}{dx}[\tan(x-y)]

We can use the chain rule, where:

(u(v(x))'=u'(v(x))\cdot v'(x)

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])

Differentiate x like normally. Implicitly differentiate for y. This yields:

=\sec^2(x-y)(1-y')

Distribute:

=\sec^2(x-y)-y'\sec^2(x-y)

And that is our left side.

Right Side:

We have:

\frac{d}{dx}[\frac{y}{8+x^2}]

We can use the quotient rule, where:

\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}

f is y. g is (8+x²). So:

=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}

Differentiate:

=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}

And that is our right side.

So, our entire equation is:

\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)

The right side cancels. Let's distribute the left:

\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2

Move -2xy to the left. So:

\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2

Factor out a y' from the right:

\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)

Divide. Therefore, dy/dx is:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}

We can factor out a (8+x²) from the denominator. So:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}

And we're done!

8 0
3 years ago
Solve for x <br> -1(-3x^2+2x+8)=1(0)
Igoryamba
Exact form: X = -4/3, 2
Decimal form: x= -1.3 repeat, 2
4 0
3 years ago
Read 2 more answers
The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

7 0
3 years ago
List the first three multiples of both 3 and 9.
tia_tia [17]

Answer:

9,18 and 27

hope this helps u

8 0
3 years ago
Read 2 more answers
Mr. Charles cut fresh roses from his garden and gave 10 roses to his neighborhood. then gave half of what was left to his niece.
Vlad [161]
14+14=28
28+10=38 roses
8 0
3 years ago
Read 2 more answers
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