Find a Solution for $y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$

• MHB
• karush
In summary, the process of solving for y using the chain rule is long, but it is possible to do it step by step.
karush
Gold Member
MHB
find y'
$$y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$$

ok this was on mml but they gave an very long process to solve it

don't see any way to expand it except recycle it via chain rule

any suggest...

I think I might square and arrange as:

$$\displaystyle y^2-7x=\sqrt{7x+\sqrt{7x}}$$

At this point you could implicitly differentiate w.r.t $$x$$ and solve for $$\displaystyle \d{y}{x}$$, or you could square again:

$$\displaystyle y^4-14xy^2+49x^2=7x+\sqrt{7x}$$

Now, implicitly differentiate w.r.t $$x$$ and solve for $$\displaystyle \d{y}{x}$$. :)

well that makes a lot more sense than redoing a chain

karush said:
find y'
$$y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$$

ok this was on mml but they gave an very long process to solve it

don't see any way to expand it except recycle it via chain rule

any suggest...
Let $u= \sqrt{7x+ \sqrt{7x}}$ so that $y= \sqrt{7x+ u}= (7x+ u)^{1/2}$. Then $\frac{dy}{dx}= \frac{1}{2}(7x+ u)^{-1/2}\left(7+ \frac{du}{dx}\right)$. To find $\frac{du}{dx}$ let $v= \sqrt{7x}$. Then $u= (7x+ v)^{1/2}$. $\frac{du}{dx}= \frac{1}{2}(7x+ v)^{-1/2}\left(7+ \frac{dv}{dx}\right)$.

$\frac{dv}{dx}= \frac{1}{2}(7x)^{-1/2}(7)$.

ok that's a very interesting process and easy to follow

but really is that the final answer

runing it thru some of the online calculators returned very long answers

symbolab returned this

$$\frac{28\sqrt{x}\sqrt{7x+\sqrt{7}\sqrt{x}}+14\sqrt{x}+\sqrt{7}}{8\sqrt{x}\sqrt{7x+\sqrt{7}\sqrt{x}}\sqrt{7x+\sqrt{7x+\sqrt{7}\sqrt{x}}}}$$

https://www.symbolab.com/solver/imp...frac{dy}{dx}, y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}

MarkFL said:
I think I might square and arrange as:

$$\displaystyle y^2-7x=\sqrt{7x+\sqrt{7x}}$$

At this point you could implicitly differentiate w.r.t $$x$$ and solve for $$\displaystyle \d{y}{x}$$, or you could square again:

$$\displaystyle y^4-14xy^2+49x^2=7x+\sqrt{7x}$$

Now, implicitly differentiate w.r.t $$x$$ and solve for $$\displaystyle \d{y}{x}$$. :)

ok I am going to do this a step at a time since I get hung up on implicit differentation

$$\displaystyle 98x-14y^2=7+\frac{\sqrt{7}}{2\sqrt{x}}$$

karush said:
ok I am going to do this a step at a time since I get hung up on implicit differentation

$$\displaystyle 98x-14y^2=7+\frac{\sqrt{7}}{2\sqrt{x}}$$

You can't treat $$y$$ as a constant, because it is a function of $$x$$. You need to use the chain rule. :)

thusly?
$\displaystyle 4y^3y'-14y^2+28xyy'+98x=7+\frac{\sqrt{7}}{2\sqrt{x}}$

karush said:
thusly?
$\displaystyle 4y^3y'-14y^2+28xyy'+98x=7+\frac{\sqrt{7}}{2\sqrt{x}}$

Check your signs on the LHS. :)

MarkFL said:
Check your signs on the LHS. :)

i suppose this$\displaystyle 4y^3y'-14y^2-28xyy'+98x=7+\frac{\sqrt{7}}{2\sqrt{x}}$if so then isolate y'

Last edited:
\begin{align*}\displaystyle 4y^3y'-14y^2-28xyy'+98x&=7+\frac{\sqrt{7}}{2\sqrt{x}}\tag{implied eq}\\ y'(4y^3y-28xy)&=7+\frac{\sqrt{7}}{2\sqrt{x}}+14y^2-98x\tag{separate variables}\\ y'&=\frac{7+\frac{\sqrt{7}}{2\sqrt{x}}+14y^2-98x}{4y^3y-28xy}\tag{divide both sides} \end{align*}

Ok, if this is correct so far the final would just be simplification which is still a complicated answer.

1. What is the solution for $y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$?

The solution for $y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$ is not a single value, but rather an infinite number of values since there are three nested square root functions. This means that there are multiple possible solutions for different values of x.

2. How do you solve $y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$?

There is no general method for solving equations with nested square roots. However, you can use trial and error, substitution, or graphing to find approximate solutions for different values of x.

3. Can this equation be simplified?

Yes, $y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$ can be simplified to $y=\sqrt{7x+\sqrt{7x\cdot\sqrt{7x}}}$. This is because the square root of a product is equal to the product of the square roots of each individual factor.

4. What are the possible values of x and y for $y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$?

The possible values of x and y for $y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$ depend on the domain of the function. If x is restricted to non-negative numbers, then y must also be non-negative. Additionally, if x is a perfect square, then y will also be a perfect square.

5. How does changing the value of x affect the value of y in $y=\sqrt{7x+\sqrt{7x+\sqrt{7x}}}$?

Changing the value of x will directly affect the value of y. As x increases, y will also increase, but at a slower rate. This is because the nested square root functions cause y to approach a maximum value as x approaches infinity. On the other hand, as x approaches zero, y will also approach zero since the innermost square root becomes negligible.

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