Find The Derivative Of The Function:$\[ Y = \sqrt{\frac{x}{x+5}} \\]$\[ Y^{\prime} = \ \square \\]

by ADMIN 99 views

Introduction

In calculus, finding the derivative of a function is a crucial concept that helps us understand the rate of change of the function with respect to its input. However, when dealing with complex functions, it can be challenging to find the derivative. In this article, we will focus on finding the derivative of the function y=xx+5y = \sqrt{\frac{x}{x+5}}. We will break down the problem into smaller steps and use various techniques to simplify the function and find its derivative.

Understanding the Function

Before we start finding the derivative, let's understand the function y=xx+5y = \sqrt{\frac{x}{x+5}}. This function is a rational function, which means it is the ratio of two polynomials. The numerator is xx, and the denominator is x+5x+5. The function is also a square root function, which means we need to take the square root of the ratio of the numerator and denominator.

Simplifying the Function

To simplify the function, we can start by rationalizing the numerator. This involves multiplying the numerator and denominator by the conjugate of the numerator. In this case, the conjugate of the numerator is x+5\sqrt{x+5}. By multiplying the numerator and denominator by x+5\sqrt{x+5}, we get:

y=xx+5β‹…x+5x+5y = \frac{x}{x+5} \cdot \frac{\sqrt{x+5}}{\sqrt{x+5}}

Simplifying the expression, we get:

y=xx+5(x+5)x+5y = \frac{x\sqrt{x+5}}{(x+5)\sqrt{x+5}}

Canceling Out the Common Factors

Now that we have simplified the function, we can cancel out the common factors in the numerator and denominator. The common factor is x+5\sqrt{x+5}, which appears in both the numerator and denominator. Canceling out the common factor, we get:

y=xx+5y = \frac{x}{x+5}

Finding the Derivative

Now that we have simplified the function, we can find its derivative. To find the derivative of the function, we can use the quotient rule, which states that if y=uvy = \frac{u}{v}, then yβ€²=vuβ€²βˆ’uvβ€²v2y^{\prime} = \frac{vu^{\prime} - uv^{\prime}}{v^2}. In this case, u=xu = x and v=x+5v = x+5. The derivative of uu is uβ€²=1u^{\prime} = 1, and the derivative of vv is vβ€²=1v^{\prime} = 1.

Applying the Quotient Rule

Using the quotient rule, we can find the derivative of the function:

yβ€²=(x+5)(1)βˆ’x(1)(x+5)2y^{\prime} = \frac{(x+5)(1) - x(1)}{(x+5)^2}

Simplifying the expression, we get:

yβ€²=x+5βˆ’x(x+5)2y^{\prime} = \frac{x+5-x}{(x+5)^2}

Simplifying the Derivative

Now that we have found the derivative, we can simplify it further. The numerator is x+5βˆ’xx+5-x, which simplifies to 55. The denominator is (x+5)2(x+5)^2, which is a squared binomial. We can simplify the squared binomial by expanding it:

(x+5)2=x2+10x+25(x+5)^2 = x^2 + 10x + 25

Final Derivative

Now that we have simplified the derivative, we can write the final derivative:

yβ€²=5x2+10x+25y^{\prime} = \frac{5}{x^2 + 10x + 25}

Conclusion

In this article, we found the derivative of the function y=xx+5y = \sqrt{\frac{x}{x+5}}. We broke down the problem into smaller steps and used various techniques to simplify the function and find its derivative. We used the quotient rule to find the derivative and simplified the expression to get the final derivative. The final derivative is yβ€²=5x2+10x+25y^{\prime} = \frac{5}{x^2 + 10x + 25}.

Future Work

In the future, we can use the derivative to analyze the behavior of the function. For example, we can use the derivative to find the critical points of the function, which are the points where the function changes from increasing to decreasing or vice versa. We can also use the derivative to find the inflection points of the function, which are the points where the function changes from concave up to concave down or vice versa.

References

  • [1] Calculus, 3rd edition, Michael Spivak
  • [2] Calculus, 2nd edition, James Stewart
  • [3] Derivatives, 2nd edition, Michael Sullivan

Glossary

  • Derivative: A measure of how a function changes as its input changes.
  • Quotient Rule: A rule for finding the derivative of a quotient of two functions.
  • Simplified: A function that has been simplified by canceling out common factors or combining like terms.
  • Squared Binomial: A binomial that has been squared, resulting in a polynomial with a squared term.

FAQs

  • Q: What is the derivative of the function y=xx+5y = \sqrt{\frac{x}{x+5}}? A: The derivative of the function is yβ€²=5x2+10x+25y^{\prime} = \frac{5}{x^2 + 10x + 25}.
  • Q: How do I find the derivative of a quotient of two functions? A: To find the derivative of a quotient of two functions, you can use the quotient rule, which states that if y=uvy = \frac{u}{v}, then yβ€²=vuβ€²βˆ’uvβ€²v2y^{\prime} = \frac{vu^{\prime} - uv^{\prime}}{v^2}.
  • Q: What is the quotient rule? A: The quotient rule is a rule for finding the derivative of a quotient of two functions. It states that if y=uvy = \frac{u}{v}, then yβ€²=vuβ€²βˆ’uvβ€²v2y^{\prime} = \frac{vu^{\prime} - uv^{\prime}}{v^2}.

Introduction

In our previous article, we found the derivative of the function y=xx+5y = \sqrt{\frac{x}{x+5}}. We broke down the problem into smaller steps and used various techniques to simplify the function and find its derivative. In this article, we will answer some of the most frequently asked questions about the derivative of a complex function.

Q: What is the derivative of the function y=xx+5y = \sqrt{\frac{x}{x+5}}?

A: The derivative of the function is yβ€²=5x2+10x+25y^{\prime} = \frac{5}{x^2 + 10x + 25}.

Q: How do I find the derivative of a quotient of two functions?

A: To find the derivative of a quotient of two functions, you can use the quotient rule, which states that if y=uvy = \frac{u}{v}, then yβ€²=vuβ€²βˆ’uvβ€²v2y^{\prime} = \frac{vu^{\prime} - uv^{\prime}}{v^2}.

Q: What is the quotient rule?

A: The quotient rule is a rule for finding the derivative of a quotient of two functions. It states that if y=uvy = \frac{u}{v}, then yβ€²=vuβ€²βˆ’uvβ€²v2y^{\prime} = \frac{vu^{\prime} - uv^{\prime}}{v^2}.

Q: How do I simplify a complex function?

A: To simplify a complex function, you can start by rationalizing the numerator. This involves multiplying the numerator and denominator by the conjugate of the numerator. You can also cancel out common factors in the numerator and denominator.

Q: What is rationalizing the numerator?

A: Rationalizing the numerator is a technique used to simplify a complex function. It involves multiplying the numerator and denominator by the conjugate of the numerator.

Q: What is the conjugate of a numerator?

A: The conjugate of a numerator is the same expression with the opposite sign. For example, the conjugate of xx is βˆ’x-x.

Q: How do I find the derivative of a function with a square root?

A: To find the derivative of a function with a square root, you can use the chain rule. The chain rule states that if y=f(g(x))y = f(g(x)), then yβ€²=fβ€²(g(x))β‹…gβ€²(x)y^{\prime} = f^{\prime}(g(x)) \cdot g^{\prime}(x).

Q: What is the chain rule?

A: The chain rule is a rule for finding the derivative of a composite function. It states that if y=f(g(x))y = f(g(x)), then yβ€²=fβ€²(g(x))β‹…gβ€²(x)y^{\prime} = f^{\prime}(g(x)) \cdot g^{\prime}(x).

Q: How do I find the derivative of a function with a squared binomial?

A: To find the derivative of a function with a squared binomial, you can use the power rule. The power rule states that if y=xny = x^n, then yβ€²=nxnβˆ’1y^{\prime} = nx^{n-1}.

Q: What is the power rule?

A: The power rule is a rule for finding the derivative of a function with a power. It states that if y=xny = x^n, then yβ€²=nxnβˆ’1y^{\prime} = nx^{n-1}.

Conclusion

In this article, we answered some of the most frequently asked questions about the derivative of a complex function. We covered topics such as the quotient rule, rationalizing the numerator, and the chain rule. We hope that this article has been helpful in understanding the derivative of a complex function.

Future Work

In the future, we can use the derivative to analyze the behavior of the function. For example, we can use the derivative to find the critical points of the function, which are the points where the function changes from increasing to decreasing or vice versa. We can also use the derivative to find the inflection points of the function, which are the points where the function changes from concave up to concave down or vice versa.

References

  • [1] Calculus, 3rd edition, Michael Spivak
  • [2] Calculus, 2nd edition, James Stewart
  • [3] Derivatives, 2nd edition, Michael Sullivan

Glossary

  • Derivative: A measure of how a function changes as its input changes.
  • Quotient Rule: A rule for finding the derivative of a quotient of two functions.
  • Rationalizing the Numerator: A technique used to simplify a complex function.
  • Conjugate: The same expression with the opposite sign.
  • Chain Rule: A rule for finding the derivative of a composite function.
  • Power Rule: A rule for finding the derivative of a function with a power.

FAQs

  • Q: What is the derivative of the function y=xx+5y = \sqrt{\frac{x}{x+5}}? A: The derivative of the function is yβ€²=5x2+10x+25y^{\prime} = \frac{5}{x^2 + 10x + 25}.
  • Q: How do I find the derivative of a quotient of two functions? A: To find the derivative of a quotient of two functions, you can use the quotient rule, which states that if y=uvy = \frac{u}{v}, then yβ€²=vuβ€²βˆ’uvβ€²v2y^{\prime} = \frac{vu^{\prime} - uv^{\prime}}{v^2}.
  • Q: What is the quotient rule? A: The quotient rule is a rule for finding the derivative of a quotient of two functions. It states that if y=uvy = \frac{u}{v}, then yβ€²=vuβ€²βˆ’uvβ€²v2y^{\prime} = \frac{vu^{\prime} - uv^{\prime}}{v^2}.

Additional Resources

  • [1] Calculus Online Course, Khan Academy
  • [2] Calculus Textbook, Michael Spivak
  • [3] Derivatives Online Course, MIT OpenCourseWare