Answer :
Answer:
Solution/answer is in the explanation.
Step-by-step explanation:
Let's check a for fun.
a)
[tex]\frac{7}{x}-\frac{1}{y}=6[/tex]
I will rewrite so I can use power rule and chain rule on the first two terms.
[tex]7x^{-1}-y^{-1}=6[/tex]
Now let's begin.
[tex]7(-1)x^{-1-1}-(-1)y^{-1-1}y'=0[/tex]
Let's simplify:
[tex]-7x^{-2}+y^{-2}y'=0[/tex]
Add [tex]7x^{-2}[/tex] on both sides:
[tex]y^{-2}y'=7x^{-2}[/tex]
Divide both sides by [tex]y^{-2}[/tex]:
[tex]y'=\frac{7x^{-2}}{y^{-2}}[/tex]
Using law of exponents we can rewrite this as:
[tex]y'=\frac{7y^2}{x^2}[/tex].
So a) does look good (Great job!).
b)
We need to solve for [tex]y[/tex] first since we want to do this explicitly. That is what that means by the way.
[tex]\frac{7}{x}-\frac{1}{y}=6[/tex]
Subtract [tex]\frac{7}{x}[/tex] on both sides:
[tex]-\frac{1}{y}=6-\frac{7}{x}[/tex]
Multiply both sides by [tex]-1[/tex]:
[tex]\frac{1}{y}=-6+\frac{7}{x}[/tex]
Multiply both sides by [tex]y[/tex]:
[tex]1=y(-6+\frac{7}{x})[/tex]
Divide both sides by [tex](-6+\frac{7}{x})[/tex]:
[tex]\frac{1}{-6+\frac{7}{x}}=y[/tex]
Symmetric property:
[tex]y=\frac{1}{-6+\frac{7}{x}}[/tex]
Multiply the right hand side be [tex]\frac{x}{x}[/tex]:
[tex]y=\frac{x}{-6x+7}[/tex] (the is the explicit equation since it is solved for [tex]y[/tex].
Now we are to differentiate this.
Let's use quotient rule [tex](\frac{f}{g})'=\frac{f'g-g'f}{g^2}[/tex]:
[tex]y'=\frac{(1)(-6x+7)-x(-6)}{(-6x+7)^2}[/tex]
[tex]y'=\frac{-6x+7+6x}{(-6x+7)^2}[/tex]
[tex]y'=\frac{7}{(-6x+7)^2}[/tex]
c) So we need to see if a) and b) are actually the same now.
For a) we go [tex]y'=\frac{7y^2}{x}[/tex] and we solve for [tex]y[/tex] in b) we go [tex]y=\frac{x}{-6x+7}[/tex].
I'm going to find [tex]y^2[/tex] so I can just replace
[tex]y=\frac{x}{-6x+7}[/tex]
Squaring both sides:
[tex]y^2=(\frac{x}{-6x+7})^2[/tex]
[tex]y^2=\frac{x^2}{(-6x+7)^2}[/tex]
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):
[tex]y'=\frac{7y^2}{x^2}[/tex]
[tex]y'=\frac{7(\frac{x^2}{(-6x+7)^2})}{x^2}[/tex]
Division by a number is the same as multiplication by the reciprocal of that number. So let's rewrite:
[tex]y'=\frac{7x^2}{x^2(-6x+7)^2}[/tex]
Now we can cancel the common factor [tex]x^2[/tex] from the top and bottom:
[tex]y'=\frac{7}{(-6x+7)^2}[/tex]
We see this is the same solution we got for b).
We have completed the objective for c).
Using differentiation, it is found that:
a)
The implicit derivative is:
[tex]y^{\prime} = \frac{7y^2}{x^2}[/tex]
b)
The explicit derivative is:
[tex]y^{\prime} = -\frac{7}{x^2}[/tex]
c)
Substituting into part a, we have that:
[tex]y^{\prime} = \frac{7}{x^2}\left[\frac{49}{x^2} - \frac{84}{x} + 36\right][/tex]
The expression given is:
[tex]\frac{7}{x} - \frac{1}{y} = 6[/tex]
Item a:
Applying implicit differentiation, we have that:
[tex]-\frac{7}{x^2}\frac{dx}{dx} + \frac{1}{y^2}\frac{dy}{dx} = 0[/tex]
[tex]\frac{1}{y^2}\frac{dy}{dx} = \frac{7}{x^2}[/tex]
[tex]\frac{dy}{dx} = \frac{7y^2}{x^2}[/tex]
Item b:
First, we solve the expression for y, then:
[tex]\frac{1}{y} = \frac{7}{x} - 6[/tex]
[tex]\frac{1}{y} = \frac{7 - 6x}{x}[/tex]
[tex]xy = 7 - 6x[/tex]
[tex]y = \frac{7 - 6x}{x}[/tex]
Applying the quotient derivative rule:
[tex]y^{\prime} = \frac{x(7-6x)^{\prime} - x^{\prime}(7-6x)}{x^2}[/tex]
[tex]y^{\prime} = \frac{-6x -7 + 6x}{x^2}[/tex]
[tex]y^{\prime} = -\frac{7}{x^2}[/tex]
Item c:
We have that the expression for the derivative is is:
[tex]y = \frac{7 - 6x}{x}[/tex]
Substituting into part a, using [tex]y = \frac{7}{x} - 6[/tex]:
[tex]y^{\prime} = \frac{7y^2}{x^2}[/tex]
[tex]y^{\prime} = \frac{7\left[\frac{7}{x} - 6]^2}{x^2}[/tex]
[tex]y^{\prime} = \frac{7}{x^2}\left[\frac{49}{x^2} - \frac{84}{x} + 36\right][/tex]
A similar problem is given at https://brainly.com/question/9543179