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

[karush]

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...

 

[MarkFL]

I think I might square and arrange as:

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

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

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

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

 

[karush]

well that makes a lot more sense than redoing a chain

 

[HOI]

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 $u= (7x+ v)^{1/2}$. $\frac{du}{dx}= \frac{1}{2}(7x+ v)^{-1/2}\left(7+ \frac{dv}{dx}\right)$.

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

 

[karush]

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}}}

 

[karush]

I think I might square and arrange as:

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

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

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

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

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

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

 

[MarkFL]

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. :)

 

[karush]

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

 

[MarkFL]

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

Check your signs on the LHS. :)

 

[karush]

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'

 

[karush]

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

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