Exercise 1.3Differentiate The Following With Respect To { X $}$:1. { Y = (x-1) \sqrt{x^2-2x+2} $}$2. { Y = \sqrt{\frac{x-1}{x+1}} $}$3. { Y = E^{\frac{1+x}{1-x}} $}$4. { Y = E^{\ln (x^2+1)} $}$5.

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In calculus, differentiating functions with respect to a variable is a fundamental concept that helps us understand the rate of change of a function. However, when dealing with complex functions, the process of differentiation can become more challenging. In this article, we will explore the differentiation of five complex functions with respect to x, using various techniques and rules of differentiation.

1. Differentiating y = (x-1) \sqrt{x^2-2x+2}

To differentiate the function y = (x-1) \sqrt{x^2-2x+2}, we will use the product rule and chain rule of differentiation.

The product rule states that if we have a function of the form y = u(x)v(x), then the derivative of y with respect to x is given by:

dy/dx = du/dx * v(x) + u(x) * dv/dx

In this case, we can rewrite the function as y = (x-1) * (x2-2x+2)(1/2).

Using the chain rule, we can differentiate the function as follows:

dy/dx = d((x-1))/dx * (x2-2x+2)(1/2) + (x-1) * d((x2-2x+2)(1/2))/dx

Now, we can differentiate the individual terms:

d((x-1))/dx = 1 d((x2-2x+2)(1/2))/dx = (1/2) * (x2-2x+2)(-1/2) * (2x-2)

Substituting these values back into the original equation, we get:

dy/dx = 1 * (x2-2x+2)(1/2) + (x-1) * (1/2) * (x2-2x+2)(-1/2) * (2x-2)

Simplifying the expression, we get:

dy/dx = (x2-2x+2)(1/2) + (x-1) * (x-1) * (x2-2x+2)(-1/2)

2. Differentiating y = \sqrt{\frac{x-1}{x+1}}

To differentiate the function y = \sqrt{\frac{x-1}{x+1}}, we will use the chain rule and quotient rule of differentiation.

The quotient rule states that if we have a function of the form y = u(x)/v(x), then the derivative of y with respect to x is given by:

dy/dx = (du/dx * v(x) - u(x) * dv/dx) / v(x)^2

In this case, we can rewrite the function as y = (x-1)^(1/2) / (x+1)^(1/2).

Using the chain rule, we can differentiate the function as follows:

dy/dx = d((x-1)^(1/2))/dx / (x+1)^(1/2) - (x-1)^(1/2) * d((x+1)^(1/2))/dx / (x+1)^(1/2)

Now, we can differentiate the individual terms:

d((x-1)^(1/2))/dx = (1/2) * (x-1)^(-1/2) * (1) d((x+1)^(1/2))/dx = (1/2) * (x+1)^(-1/2) * (1)

Substituting these values back into the original equation, we get:

dy/dx = (1/2) * (x-1)^(-1/2) / (x+1)^(1/2) - (x-1)^(1/2) * (1/2) * (x+1)^(-1/2) / (x+1)^(1/2)

Simplifying the expression, we get:

dy/dx = (1/2) * (x-1)^(-1/2) / (x+1)^(1/2) - (1/2) * (x-1)^(1/2) / (x+1)^(1/2)

3. Differentiating y = e^{\frac{1+x}{1-x}}

To differentiate the function y = e^{\frac{1+x}{1-x}}, we will use the chain rule and quotient rule of differentiation.

The quotient rule states that if we have a function of the form y = u(x)/v(x), then the derivative of y with respect to x is given by:

dy/dx = (du/dx * v(x) - u(x) * dv/dx) / v(x)^2

In this case, we can rewrite the function as y = e^((1+x)/(1-x)).

Using the chain rule, we can differentiate the function as follows:

dy/dx = d(e^((1+x)/(1-x)))/dx

Now, we can differentiate the individual terms:

d(e^((1+x)/(1-x)))/dx = e^((1+x)/(1-x)) * d((1+x)/(1-x))/dx

Using the quotient rule, we can differentiate the individual terms:

d((1+x)/(1-x))/dx = (d(1+x)/dx * (1-x) - (1+x) * d(1-x)/dx) / (1-x)^2

Now, we can differentiate the individual terms:

d(1+x)/dx = 1 d(1-x)/dx = -1

Substituting these values back into the original equation, we get:

d((1+x)/(1-x))/dx = (1 * (1-x) - (1+x) * (-1)) / (1-x)^2

Simplifying the expression, we get:

d((1+x)/(1-x))/dx = (1-x + 1 + x) / (1-x)^2

d((1+x)/(1-x))/dx = 2 / (1-x)^2

Substituting this value back into the original equation, we get:

dy/dx = e^((1+x)/(1-x)) * 2 / (1-x)^2

4. Differentiating y = e^{\ln (x^2+1)}

To differentiate the function y = e^{\ln (x^2+1)}, we will use the chain rule and logarithmic rule of differentiation.

The logarithmic rule states that if we have a function of the form y = ln(u(x)), then the derivative of y with respect to x is given by:

dy/dx = (1/u(x)) * du/dx

In this case, we can rewrite the function as y = e(ln(x2+1)).

Using the chain rule, we can differentiate the function as follows:

dy/dx = d(e(ln(x2+1)))/dx

Now, we can differentiate the individual terms:

d(e(ln(x2+1)))/dx = e(ln(x2+1)) * d(ln(x^2+1))/dx

Using the logarithmic rule, we can differentiate the individual terms:

d(ln(x^2+1))/dx = (1/(x^2+1)) * d(x^2+1)/dx

Now, we can differentiate the individual terms:

d(x^2+1)/dx = 2x

Substituting this value back into the original equation, we get:

d(ln(x^2+1))/dx = (1/(x^2+1)) * 2x

Simplifying the expression, we get:

d(ln(x^2+1))/dx = 2x / (x^2+1)

Substituting this value back into the original equation, we get:

dy/dx = e(ln(x2+1)) * 2x / (x^2+1)

Simplifying the expression, we get:

dy/dx = x^2+1 * 2x / (x^2+1)

dy/dx = 2x

Conclusion

In this article, we have explored the differentiation of five complex functions with respect to x, using various techniques and rules of differentiation. We have seen how to use the product rule, chain rule, quotient rule, and logarithmic rule to differentiate functions with multiple terms and complex expressions. By applying these rules and techniques, we can differentiate a wide range of functions and gain a deeper understanding of the behavior of these functions.

References

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

Further Reading

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

  • [1] Khan Academy: Calculus
  • [2] MIT OpenCourseWare: Calculus
  • [3] Wolfram Alpha: Calculus

In our previous article, we explored the differentiation of five complex functions with respect to x, using various techniques and rules of differentiation. In this article, we will answer some common questions related to differentiation of complex functions.

Q: What is the product rule of differentiation?

A: The product rule of differentiation states that if we have a function of the form y = u(x)v(x), then the derivative of y with respect to x is given by:

dy/dx = du/dx * v(x) + u(x) * dv/dx

Q: What is the chain rule of differentiation?

A: The chain rule of differentiation states that if we have a function of the form y = f(u(x)), then the derivative of y with respect to x is given by:

dy/dx = df/du * du/dx

Q: What is the quotient rule of differentiation?

A: The quotient rule of differentiation states that if we have a function of the form y = u(x)/v(x), then the derivative of y with respect to x is given by:

dy/dx = (du/dx * v(x) - u(x) * dv/dx) / v(x)^2

Q: How do I differentiate a function with multiple terms?

A: To differentiate a function with multiple terms, you can use the product rule and chain rule of differentiation. For example, if we have a function of the form y = u(x)v(x) + w(x), then the derivative of y with respect to x is given by:

dy/dx = du/dx * v(x) + u(x) * dv/dx + dw/dx

Q: How do I differentiate a function with a complex expression?

A: To differentiate a function with a complex expression, you can use the chain rule and logarithmic rule of differentiation. For example, if we have a function of the form y = e(ln(x2+1)), then the derivative of y with respect to x is given by:

dy/dx = e(ln(x2+1)) * d(ln(x^2+1))/dx

Using the logarithmic rule, we can differentiate the individual terms:

d(ln(x^2+1))/dx = (1/(x^2+1)) * d(x^2+1)/dx

Now, we can differentiate the individual terms:

d(x^2+1)/dx = 2x

Substituting this value back into the original equation, we get:

d(ln(x^2+1))/dx = (1/(x^2+1)) * 2x

Simplifying the expression, we get:

d(ln(x^2+1))/dx = 2x / (x^2+1)

Substituting this value back into the original equation, we get:

dy/dx = e(ln(x2+1)) * 2x / (x^2+1)

Simplifying the expression, we get:

dy/dx = x^2+1 * 2x / (x^2+1)

dy/dx = 2x

Q: What are some common mistakes to avoid when differentiating complex functions?

A: Some common mistakes to avoid when differentiating complex functions include:

  • Forgetting to use the chain rule and logarithmic rule of differentiation
  • Not simplifying the expression before differentiating
  • Not checking the domain of the function before differentiating
  • Not using the correct rules of differentiation for the given function

Q: How can I practice differentiating complex functions?

A: You can practice differentiating complex functions by:

  • Working through examples and exercises in a calculus textbook
  • Using online resources and calculators to check your work
  • Practicing differentiating functions with multiple terms and complex expressions
  • Working with a tutor or study group to get help and feedback

By following these tips and practicing regularly, you can become proficient in differentiating complex functions and gain a deeper understanding of calculus.