S8.2.6.3 differenciate by separation and power rule

In summary, to find y' for the equation $\sqrt{x}+\sqrt{y}=1$, we first isolate y by squaring both sides, resulting in $y=(1-\sqrt{x})^2$. Then, we differentiate both sides, simplify, and rationalize the denominator, giving us $y'=-\dfrac{1-\sqrt{x}}{\sqrt{x}}$. This can also be written as $y'=\dfrac{\sqrt{x}-1}{\sqrt{x}}$, which appears to be a "magic wand" solution.
  • #1
karush
Gold Member
MHB
3,269
5
$\tiny{s8.2.6.3}$
Find y' $\sqrt{x}+\sqrt{y}=1$
\begin{array}{lll}
\textit{isolate }y
&\sqrt{y}=1-\sqrt{x}
&(1)\\ \\
\textit{square both sides}
&y=(1-\sqrt{x})^2
&(2)\\ \\
\textit{differentiate both sides}
&y'=2\left(1-\sqrt{x}\right)\left(-\dfrac{1}{2\sqrt{x}}\right)&
(3)\\ \\
\textit{simplify}
&y'=-\dfrac{1-\sqrt{x}}{\sqrt{x}}
&(4)
\end{array}

well we could rationalize the denominator but why?
hopefully correct
 
Physics news on Phys.org
  • #2
implicitly ...

$\dfrac{1}{2\sqrt{x}} + \dfrac{y’}{2\sqrt{y}} = 0$

$y’ = -\dfrac{\sqrt{y}}{\sqrt{x}} = \dfrac{\sqrt{x}-1}{\sqrt{x}}$
 
  • #3
skeeter said:
implicitly ...

$\dfrac{1}{2\sqrt{x}} + \dfrac{y’}{2\sqrt{y}} = 0$

$y’ = -\dfrac{\sqrt{y}}{\sqrt{x}} = \dfrac{\sqrt{x}-1}{\sqrt{x}}$
that looks like magic wand stuff 😎

mahalo tho,,,
 

FAQ: S8.2.6.3 differenciate by separation and power rule

What is the separation rule in calculus?

The separation rule, also known as the constant multiple rule, states that when differentiating a function multiplied by a constant, the constant can be pulled out and multiplied by the derivative of the function. In other words, if f(x) is a function and c is a constant, then the derivative of c*f(x) is equal to c times the derivative of f(x).

What is the power rule in calculus?

The power rule is a basic rule in calculus that allows us to find the derivative of a function raised to a power. Specifically, if f(x) is a function and n is any real number, then the derivative of f(x)^n is equal to n times f(x)^(n-1) times the derivative of f(x).

How do you use the separation and power rules together?

When differentiating a function that is a product of two functions raised to a power, we can use both the separation rule and the power rule. First, we use the separation rule to pull out any constants in front of the function. Then, we use the power rule to find the derivative of each function raised to a power. Finally, we multiply these two derivatives together to get the final result.

Can the separation and power rules be used for any type of function?

No, the separation and power rules can only be used for functions that are in the form of a constant times a function raised to a power. If a function is more complex, such as a sum or difference of functions, then we need to use other differentiation rules.

Why are the separation and power rules important in calculus?

The separation and power rules are important because they allow us to easily find the derivative of a function that is a product of two functions raised to a power. This is a common type of function that appears in many real-world applications, so understanding these rules is crucial for solving problems in fields such as physics, engineering, and economics.

Similar threads

MHB Can Implicit Differentiation Be Done by Separation of Variables?
Replies
2
Views
972
MHB 48.2.2.23 IVP and min valuew
Replies
3
Views
918
MHB -apc.2.1.06 crosses the x-axis at one point in the interval [0,1]
Replies
4
Views
1K
MHB Evaluate S8.4.3.84 Integral from 1 to 3
Replies
3
Views
1K
MHB What Is the Value of the Integral $\int_0^1 xe^x \, dx$?
Replies
1
Views
976
MHB 2.1.312 AP Calculus Exam Int of half circle
Replies
5
Views
1K
MHB Apc.2.8.1 ap vertical circular cylinder related rates
Replies
2
Views
2K
MHB 4.1.237 AP calculus exam find area
Replies
4
Views
2K
MHB 7.3.5 Integral with trig substitution
Replies
1
Views
1K
MHB 206.08.04.59 int completing the square
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top