Explore IDNLearn.com's extensive Q&A database and find the answers you need. Get prompt and accurate answers to your questions from our community of knowledgeable experts.
Sagot :
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]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.
