Answer :
To prove that if [tex]\( y = \left(x + \sqrt{x^2 - 1}\right)^m \)[/tex], then [tex]\[ \left(x^2 - 1\right) \left(\frac{d y}{d x}\right)^2 = m^2 y^2, \][/tex]
we can follow these steps:
1. Define the function [tex]\( y \)[/tex]:
[tex]\[ y = \left(x + \sqrt{x^2 - 1}\right)^m \][/tex]
2. Compute the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
Using the chain rule, we find [tex]\( \frac{d y}{d x} \)[/tex].
Let [tex]\( u = x + \sqrt{x^2 - 1} \)[/tex], then [tex]\( y = u^m \)[/tex].
First, find [tex]\( \frac{d u}{d x} \)[/tex]:
[tex]\[ \frac{d u}{d x} = 1 + \frac{d}{d x} \left( \sqrt{x^2 - 1} \right) \][/tex]
Using the chain rule for the square root:
[tex]\[ \frac{d}{d x} \left( \sqrt{x^2 - 1} \right) = \frac{1}{2} \cdot \left( x^2 - 1 \right)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}} \][/tex]
Thus:
[tex]\[ \frac{d u}{d x} = 1 + \frac{x}{\sqrt{x^2 - 1}} \][/tex]
Now, use the chain rule to find [tex]\( \frac{d y}{d x} \)[/tex]:
[tex]\[ \frac{d y}{d x} = \frac{d}{d x} \left( u^m \right) = m \cdot u^{m-1} \cdot \frac{d u}{d x} \][/tex]
Substituting [tex]\( u \)[/tex] back:
[tex]\[ \frac{d y}{d x} = m \left( x + \sqrt{x^2 - 1} \right)^{m-1} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right) \][/tex]
3. Compute [tex]\(\left(\frac{d y}{d x}\right)^2 \)[/tex]:
[tex]\[ \left(\frac{d y}{d x}\right)^2 = \left[ m \left( x + \sqrt{x^2 - 1} \right)^{m-1} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right) \right]^2 \][/tex]
[tex]\[ = m^2 \left( x + \sqrt{x^2 - 1} \right)^{2(m-1)} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right)^2 \][/tex]
4. Multiply by [tex]\( x^2 - 1 \)[/tex]:
We need to show that:
[tex]\[ \left(x^2 - 1\right)\left(\frac{d y}{d x}\right)^2 = m^2 y^2 \][/tex]
Substitute [tex]\( \left(\frac{d y}{d x}\right)^2 \)[/tex]:
[tex]\[ (x^2 - 1) \left[\left(\frac{d y}{d x}\right)^2\right] \][/tex]
[tex]\[ = (x^2 - 1) \left[ m^2 \left( x + \sqrt{x^2 - 1} \right)^{2(m-1)} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right)^2 \right] \][/tex]
5. Simplify the expression:
Note that:
[tex]\[ y^2 = \left( \left( x + \sqrt{x^2 - 1} \right)^m \right)^2 = \left( x + \sqrt{x^2 - 1} \right)^{2m} \][/tex]
For the term containing [tex]\( (x + \sqrt{x^2 - 1})^{2(m-1)} \)[/tex]:
[tex]\[ (x^2 - 1) m^2 \left( x + \sqrt{x^2 - 1} \right)^{2(m-1)} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right)^2 \][/tex]
Multiply [tex]\( x^2 - 1 \)[/tex] by [tex]\( \left(1 + \frac{x}{\sqrt{x^2 - 1}}\right)^2 \)[/tex]:
[tex]\[ (x^2 - 1) \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right)^2 = (x^2 - 1) \left( 1 + \frac{2x}{\sqrt{x^2 - 1}} + \frac{x^2}{x^2 - 1} \right) = x^2 - 1 + 2x \sqrt{x^2 - 1} + x^2 = (x + \sqrt{x^2 - 1})^2 \][/tex]
Thus:
[tex]\[ \left(x^2 - 1\right) \left(\frac{d y}{d x}\right)^2 = m^2 (x + \sqrt{x^2 - 1})^{2(m-1)} \left( (x + \sqrt{x^2 - 1})^2 \right) \][/tex]
[tex]\[ = m^2 (x + \sqrt{x^2 - 1})^{2m} = m^2 y^2 \][/tex]
Therefore, we have successfully proved that:
[tex]\[ \left(x^2 - 1\right)\left(\frac{d y}{d x}\right)^2 = m^2 y^2 \][/tex]
we can follow these steps:
1. Define the function [tex]\( y \)[/tex]:
[tex]\[ y = \left(x + \sqrt{x^2 - 1}\right)^m \][/tex]
2. Compute the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
Using the chain rule, we find [tex]\( \frac{d y}{d x} \)[/tex].
Let [tex]\( u = x + \sqrt{x^2 - 1} \)[/tex], then [tex]\( y = u^m \)[/tex].
First, find [tex]\( \frac{d u}{d x} \)[/tex]:
[tex]\[ \frac{d u}{d x} = 1 + \frac{d}{d x} \left( \sqrt{x^2 - 1} \right) \][/tex]
Using the chain rule for the square root:
[tex]\[ \frac{d}{d x} \left( \sqrt{x^2 - 1} \right) = \frac{1}{2} \cdot \left( x^2 - 1 \right)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}} \][/tex]
Thus:
[tex]\[ \frac{d u}{d x} = 1 + \frac{x}{\sqrt{x^2 - 1}} \][/tex]
Now, use the chain rule to find [tex]\( \frac{d y}{d x} \)[/tex]:
[tex]\[ \frac{d y}{d x} = \frac{d}{d x} \left( u^m \right) = m \cdot u^{m-1} \cdot \frac{d u}{d x} \][/tex]
Substituting [tex]\( u \)[/tex] back:
[tex]\[ \frac{d y}{d x} = m \left( x + \sqrt{x^2 - 1} \right)^{m-1} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right) \][/tex]
3. Compute [tex]\(\left(\frac{d y}{d x}\right)^2 \)[/tex]:
[tex]\[ \left(\frac{d y}{d x}\right)^2 = \left[ m \left( x + \sqrt{x^2 - 1} \right)^{m-1} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right) \right]^2 \][/tex]
[tex]\[ = m^2 \left( x + \sqrt{x^2 - 1} \right)^{2(m-1)} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right)^2 \][/tex]
4. Multiply by [tex]\( x^2 - 1 \)[/tex]:
We need to show that:
[tex]\[ \left(x^2 - 1\right)\left(\frac{d y}{d x}\right)^2 = m^2 y^2 \][/tex]
Substitute [tex]\( \left(\frac{d y}{d x}\right)^2 \)[/tex]:
[tex]\[ (x^2 - 1) \left[\left(\frac{d y}{d x}\right)^2\right] \][/tex]
[tex]\[ = (x^2 - 1) \left[ m^2 \left( x + \sqrt{x^2 - 1} \right)^{2(m-1)} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right)^2 \right] \][/tex]
5. Simplify the expression:
Note that:
[tex]\[ y^2 = \left( \left( x + \sqrt{x^2 - 1} \right)^m \right)^2 = \left( x + \sqrt{x^2 - 1} \right)^{2m} \][/tex]
For the term containing [tex]\( (x + \sqrt{x^2 - 1})^{2(m-1)} \)[/tex]:
[tex]\[ (x^2 - 1) m^2 \left( x + \sqrt{x^2 - 1} \right)^{2(m-1)} \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right)^2 \][/tex]
Multiply [tex]\( x^2 - 1 \)[/tex] by [tex]\( \left(1 + \frac{x}{\sqrt{x^2 - 1}}\right)^2 \)[/tex]:
[tex]\[ (x^2 - 1) \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right)^2 = (x^2 - 1) \left( 1 + \frac{2x}{\sqrt{x^2 - 1}} + \frac{x^2}{x^2 - 1} \right) = x^2 - 1 + 2x \sqrt{x^2 - 1} + x^2 = (x + \sqrt{x^2 - 1})^2 \][/tex]
Thus:
[tex]\[ \left(x^2 - 1\right) \left(\frac{d y}{d x}\right)^2 = m^2 (x + \sqrt{x^2 - 1})^{2(m-1)} \left( (x + \sqrt{x^2 - 1})^2 \right) \][/tex]
[tex]\[ = m^2 (x + \sqrt{x^2 - 1})^{2m} = m^2 y^2 \][/tex]
Therefore, we have successfully proved that:
[tex]\[ \left(x^2 - 1\right)\left(\frac{d y}{d x}\right)^2 = m^2 y^2 \][/tex]
