Given That $g(x)=\frac{1}{x^n}$, Show That $x G^{\prime}(x)+n G(x)=0$.

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In this article, we will delve into the world of calculus and explore a fundamental identity involving a given function. The function in question is defined as $g(x)=\frac{1}{x^n}$, and we aim to demonstrate that $x g^{\prime}(x)+n g(x)=0$. This identity is crucial in various mathematical contexts, and understanding its derivation will provide valuable insights into the properties of the given function.

Understanding the Given Function

The function $g(x)=\frac{1}{x^n}$ is a rational function, where the numerator is a constant (1) and the denominator is a variable raised to a power of nn. This function is defined for all real values of xx except when x=0x=0, as division by zero is undefined.

Deriving the Derivative of the Function

To derive the derivative of the function $g(x)=\frac{1}{x^n}$, we will employ the power rule of differentiation. The power rule states that if f(x)=xnf(x)=x^n, then f′(x)=nxn−1f^{\prime}(x)=n x^{n-1}. We can apply this rule to our function by recognizing that $g(x)=x^{-n}$.

Using the power rule, we can write the derivative of g(x)g(x) as:

g′(x)=ddx(x−n)=−nx−n−1g^{\prime}(x) = \frac{d}{dx} \left( x^{-n} \right) = -n x^{-n-1}

Deriving the Key Identity

Now that we have the derivative of the function, we can proceed to derive the key identity. We are given the expression $x g^{\prime}(x)+n g(x)$, and we aim to show that it equals zero.

Substituting the expression for g′(x)g^{\prime}(x) that we derived earlier, we get:

xg′(x)+ng(x)=x(−nx−n−1)+n(x−n)x g^{\prime}(x)+n g(x) = x \left( -n x^{-n-1} \right) + n \left( x^{-n} \right)

Simplifying the expression, we get:

xg′(x)+ng(x)=−nx−n+nx−nx g^{\prime}(x)+n g(x) = -n x^{-n} + n x^{-n}

Notice that the two terms on the right-hand side are identical, except for the sign. Therefore, they cancel each other out, leaving us with:

xg′(x)+ng(x)=0x g^{\prime}(x)+n g(x) = 0

Conclusion

In this article, we have successfully derived the key identity $x g^{\prime}(x)+n g(x)=0$ for the given function $g(x)=\frac{1}{x^n}$. This identity is a fundamental result in calculus, and understanding its derivation provides valuable insights into the properties of the given function. We hope that this article has been informative and helpful in your mathematical journey.

Additional Resources

For further reading on calculus and mathematical analysis, we recommend the following resources:

Final Thoughts

In our previous article, we derived the key identity $x g^{\prime}(x)+n g(x)=0$ for the given function $g(x)=\frac{1}{x^n}$. However, we understand that some readers may still have questions or doubts about the derivation. In this article, we will address some of the most frequently asked questions related to the derivation of the key identity.

Q: What is the significance of the power rule of differentiation?

A: The power rule of differentiation is a fundamental concept in calculus that states that if f(x)=xnf(x)=x^n, then f′(x)=nxn−1f^{\prime}(x)=n x^{n-1}. This rule is essential for deriving the derivative of the given function $g(x)=\frac{1}{x^n}$.

Q: Why is the function $g(x)=\frac{1}{x^n}$ defined for all real values of xx except when x=0x=0?

A: The function $g(x)=\frac{1}{x^n}$ is defined for all real values of xx except when x=0x=0 because division by zero is undefined. When x=0x=0, the denominator of the function becomes zero, and the function is not defined.

Q: How do we apply the power rule of differentiation to the function $g(x)=\frac{1}{x^n}$?

A: To apply the power rule of differentiation to the function $g(x)=\frac{1}{x^n}$, we recognize that $g(x)=x^{-n}$. Then, we can use the power rule to write the derivative of g(x)g(x) as:

g′(x)=ddx(x−n)=−nx−n−1g^{\prime}(x) = \frac{d}{dx} \left( x^{-n} \right) = -n x^{-n-1}

Q: Why do the two terms on the right-hand side of the expression $x g^{\prime}(x)+n g(x)$ cancel each other out?

A: The two terms on the right-hand side of the expression $x g^{\prime}(x)+n g(x)$ cancel each other out because they are identical, except for the sign. When we simplify the expression, we get:

xg′(x)+ng(x)=−nx−n+nx−nx g^{\prime}(x)+n g(x) = -n x^{-n} + n x^{-n}

Since the two terms are identical, they cancel each other out, leaving us with:

xg′(x)+ng(x)=0x g^{\prime}(x)+n g(x) = 0

Q: What are some real-world applications of the key identity $x g^{\prime}(x)+n g(x)=0$?

A: The key identity $x g^{\prime}(x)+n g(x)=0$ has numerous real-world applications in various fields, including physics, engineering, and economics. For example, it can be used to model population growth, chemical reactions, and economic systems.

Q: How can I apply the key identity $x g^{\prime}(x)+n g(x)=0$ to solve problems in calculus?

A: To apply the key identity $x g^{\prime}(x)+n g(x)=0$ to solve problems in calculus, you can use it as a tool to derive new identities and equations. For example, you can use it to derive the derivative of a function or to solve optimization problems.

Conclusion

In this article, we have addressed some of the most frequently asked questions related to the derivation of the key identity $x g^{\prime}(x)+n g(x)=0$. We hope that this article has been informative and helpful in your mathematical journey. If you have any further questions or doubts, please don't hesitate to ask.

Additional Resources

For further reading on calculus and mathematical analysis, we recommend the following resources:

Final Thoughts

In conclusion, the key identity $x g^{\prime}(x)+n g(x)=0$ is a fundamental result in calculus that has numerous real-world applications. Understanding its derivation provides valuable insights into the properties of the given function and is essential for further mathematical analysis. We hope that this article has been informative and helpful in your mathematical journey.