Mathematics, 02.10.2020 14:01 stheo

Consider the following. 7/x - 1/y = 6
(a) I solved part a for y' using implicit differentiation, it is y'= (7y²) / (x²)
(b) Solve the equation explicitly for y and differentiate to get y' in terms of x.
(c) Check the consistency of your solutions to part a and b by substituting the expression for y into your solution for part a

B and C are the only ones i'm having trouble with. I know how to do this in general, just that the fractions in this particular case are messing me up, if you could help me that would be much appreciated!

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Answer from: bhjbh7ubb

Using differentiation, it is found that:

The implicit derivative is:

$y^{\prime} = \frac{7y^2}{x^2}$

The explicit derivative is:

$y^{\prime} = -\frac{7}{x^2}$

Substituting into part a, we have that:

$y^{\prime} = \frac{7}{x^2}\left[\frac{49}{x^2} - \frac{84}{x} + 36\right]$

The expression given is:

$\frac{7}{x} - \frac{1}{y} = 6$

Item a:

Applying implicit differentiation, we have that:

$-\frac{7}{x^2}\frac{dx}{dx} + \frac{1}{y^2}\frac{dy}{dx} = 0$

$\frac{1}{y^2}\frac{dy}{dx} = \frac{7}{x^2}$

$\frac{dy}{dx} = \frac{7y^2}{x^2}$

Item b:

First, we solve the expression for y, then:

$\frac{1}{y} = \frac{7}{x} - 6$

$\frac{1}{y} = \frac{7 - 6x}{x}$

xy = 7 - 6x

$y = \frac{7 - 6x}{x}$

Applying the quotient derivative rule:

$y^{\prime} = \frac{x(7-6x)^{\prime} - x^{\prime}(7-6x)}{x^2}$

$y^{\prime} = \frac{-6x -7 + 6x}{x^2}$

$y^{\prime} = -\frac{7}{x^2}$

Item c:

We have that the expression for the derivative is is:

$y = \frac{7 - 6x}{x}$

Substituting into part a, using $y = \frac{7}{x} - 6$ :

$y^{\prime} = \frac{7y^2}{x^2}$

$y^{\prime} = \frac{7\left[\frac{7}{x} - 6]^2}{x^2}$

$y^{\prime} = \frac{7}{x^2}\left[\frac{49}{x^2} - \frac{84}{x} + 36\right]$

A similar problem is given at link

Answer from: alex7881

Solution/answer is in the explanation.

Step-by-step explanation:

Let's check a for fun.

$\frac{7}{x}-\frac{1}{y}=6$

I will rewrite so I can use power rule and chain rule on the first two terms.

$7x^{-1}-y^{-1}=6$

Now let's begin.

$7(-1)x^{-1-1}-(-1)y^{-1-1}y'=0$

Let's simplify:

$-7x^{-2}+y^{-2}y'=0$

Add $7x^{-2}$ on both sides:

$y^{-2}y'=7x^{-2}$

Divide both sides by $y^{-2}$ :

$y'=\frac{7x^{-2}}{y^{-2}}$

Using law of exponents we can rewrite this as:

$y'=\frac{7y^2}{x^2}$ .

So a) does look good (Great job!).

We need to solve for first since we want to do this explicitly. That is what that means by the way.

$\frac{7}{x}-\frac{1}{y}=6$

Subtract $\frac{7}{x}$ on both sides:

$-\frac{1}{y}=6-\frac{7}{x}$

Multiply both sides by :

$\frac{1}{y}=-6+\frac{7}{x}$

Multiply both sides by :

$1=y(-6+\frac{7}{x})$

Divide both sides by $(-6+\frac{7}{x})$ :

$\frac{1}{-6+\frac{7}{x}}=y$

Symmetric property:

$y=\frac{1}{-6+\frac{7}{x}}$

Multiply the right hand side be $\frac{x}{x}$ :

$y=\frac{x}{-6x+7}$ (the is the explicit equation since it is solved for .

Now we are to differentiate this.

Let's use quotient rule $(\frac{f}{g})'=\frac{f'g-g'f}{g^2}$ :

$y'=\frac{(1)(-6x+7)-x(-6)}{(-6x+7)^2}$

$y'=\frac{-6x+7+6x}{(-6x+7)^2}$

$y'=\frac{7}{(-6x+7)^2}$

c) So we need to see if a) and b) are actually the same now.

For a) we go $y'=\frac{7y^2}{x}$ and we solve for in b) we go $y=\frac{x}{-6x+7}$ .

I'm going to find y^2 so I can just replace

$y=\frac{x}{-6x+7}$

Squaring both sides:

$y^2=(\frac{x}{-6x+7})^2$

$y^2=\frac{x^2}{(-6x+7)^2}$

I will leave it as this because we may not have to do further work such as expanding the bottom.

Let's plug this into our solution for a):

$y'=\frac{7y^2}{x^2}$

$y'=\frac{7(\frac{x^2}{(-6x+7)^2})}{x^2}$

Division by a number is the same as multiplication by the reciprocal of that number. So let's rewrite:

$y'=\frac{7x^2}{x^2(-6x+7)^2}$

Now we can cancel the common factor x^2 from the top and bottom:

$y'=\frac{7}{(-6x+7)^2}$

We see this is the same solution we got for b).

We have completed the objective for c).

Answer from: Quest

step-by-step explanation:

10 ptsfind the area of the inside of the frame where the picture would belong. then find the area of

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use context clues to figure out the meaning of the bolded word:

the sage gardener knew when she had to water her plants.

question 8 options:

mint plant

wise

happy

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Consider the following. 7/x - 1/y = 6
(a) I solved part a for y' using implicit differentiati...

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