Finding Dy/dx Derivatives A Comprehensive Guide With Examples

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In calculus, finding the derivative, denoted as dy/dx, is a fundamental operation. It represents the instantaneous rate of change of a function y with respect to its variable x. This article provides a comprehensive guide on how to calculate dy/dx for various functions and expressions, complete with step-by-step explanations and examples. Whether you're dealing with simple algebraic functions, trigonometric functions, or complex expressions, this guide will equip you with the necessary tools and techniques.

1. Derivative of y = √x

To find the derivative of y = √x, we first express the square root as a power of x, which is y = x^(1/2). We will leverage the power rule for differentiation, which states that if y = x^n, then dy/dx = nx^(n-1). Applying this rule, we differentiate the function step by step.

Step-by-Step Differentiation

  1. Rewrite the function: Start by rewriting y = √x as y = x^(1/2). This conversion makes it easier to apply the power rule directly.
  2. Apply the power rule: Using the power rule, where n = 1/2, we get dy/dx = (1/2)x^((1/2) - 1). This step involves multiplying the function by the exponent and reducing the exponent by 1.
  3. Simplify the exponent: Simplify the exponent (1/2) - 1 to -1/2, so the derivative becomes dy/dx = (1/2)x^(-1/2). This simplification prepares the expression for the final form.
  4. Rewrite with a positive exponent: To express the derivative with a positive exponent, we move x^(-1/2) to the denominator as x^(1/2). This gives us dy/dx = 1/(2x^(1/2)). Simplifying exponents helps in better interpretation and further calculations.
  5. Convert back to radical form: Finally, convert x^(1/2) back to its radical form, which is √x. Therefore, the derivative is dy/dx = 1/(2√x). This final conversion provides a clear and concise representation of the derivative, suitable for various applications and interpretations.

Thus, the derivative of y = √x is dy/dx = 1/(2√x). This result is a cornerstone in calculus and is frequently used in various mathematical and scientific contexts. The power rule's application here exemplifies its utility in differentiating algebraic functions, making it an essential technique for calculus students and practitioners alike.

2. Derivative of y = cos(√q)

To find the derivative of y = cos(√q), we must specify the variable with respect to which we are differentiating. Assuming we want to find dy/dq, we will use the chain rule. The chain rule is crucial for differentiating composite functions, which are functions within functions. In this case, we have the cosine function applied to the square root of q.

Step-by-Step Differentiation

  1. Identify the composite functions: Recognize that y = cos(√q) is a composite function. Let u = √q and y = cos(u). This decomposition helps in applying the chain rule effectively. Identifying inner and outer functions is the first step in using the chain rule.
  2. Find the derivative of the outer function with respect to the inner function: Differentiate y = cos(u) with respect to u. The derivative of cos(u) is -sin(u), so dy/du = -sin(u). This step involves knowing the basic derivatives of trigonometric functions.
  3. Find the derivative of the inner function with respect to q: Differentiate u = √q with respect to q. As we found in the first example, the derivative of √q or q^(1/2) is (1/2)q^(-1/2), which can be rewritten as 1/(2√q). Thus, du/dq = 1/(2√q). Differentiating the inner function is essential for the chain rule's correct application.
  4. Apply the chain rule: The chain rule states that dy/dq = (dy/du) * (du/dq). Substituting the derivatives we found, we get dy/dq = -sin(u) * (1/(2√q)). The chain rule links the rates of change of the composite functions.
  5. Substitute back for u: Replace u with √q in the expression. This gives us dy/dq = -sin(√q) * (1/(2√q)). Substituting back ensures the final derivative is in terms of the original variable.
  6. Simplify the expression: Simplify the expression to obtain the final derivative: dy/dq = -sin(√q) / (2√q). This simplified form is the derivative of the original function.

Therefore, the derivative of y = cos(√q) with respect to q is dy/dq = -sin(√q) / (2√q). The chain rule is fundamental in calculus for handling such composite functions, ensuring accurate differentiation. This example showcases its importance and application in finding derivatives of complex functions.

3. Derivative of (x sin x) / cos x

To find the derivative of (x sin x) / cos x, we can simplify the function first and then apply differentiation rules. Simplifying the expression makes it easier to differentiate, especially when dealing with trigonometric functions. Let's rewrite the function using trigonometric identities and then differentiate.

Simplifying the Function

  1. Rewrite the function: Observe that sin(x) / cos(x) is equivalent to tan(x). Thus, we can rewrite the function as y = x tan(x). This simplification reduces the complexity of the expression and sets the stage for applying the product rule.

Step-by-Step Differentiation

  1. Apply the product rule: The product rule states that if y = uv, then dy/dx = u'(v) + v'(u). In this case, let u = x and v = tan(x). The product rule is essential for differentiating the product of two functions.
  2. Find the derivatives of u and v: The derivative of u = x with respect to x is u' = 1. The derivative of v = tan(x) with respect to x is v' = sec^2(x). Knowing basic derivatives is crucial for applying the product rule.
  3. Apply the product rule formula: Substitute the derivatives into the product rule formula: dy/dx = (1) * tan(x) + sec^2(x) * (x). This step involves plugging in the derived values into the product rule formula.
  4. Simplify the expression: Simplify the expression to get dy/dx = tan(x) + x sec^2(x). This simplified form represents the derivative of the original function.

Thus, the derivative of y = (x sin x) / cos x is dy/dx = tan(x) + x sec^2(x). The simplification using trigonometric identities followed by the application of the product rule demonstrates an effective method for differentiating such functions. This approach highlights the importance of recognizing opportunities to simplify expressions before differentiation, making the process more manageable and accurate.

4. Derivative of x² + sin(x)^cos(x)

To find the derivative of y = x² + sin(x)^cos(x), we need to differentiate each term separately. The first term, x², is straightforward using the power rule. The second term, sin(x)^cos(x), requires a more complex approach involving logarithmic differentiation. This method is particularly useful when the variable appears in both the base and the exponent of a function.

Step-by-Step Differentiation for x²

  1. Apply the power rule: For y = x², the power rule states that dy/dx = nx^(n-1). Applying this, we get dy/dx = 2x^(2-1) = 2x. This step is a direct application of the power rule, a fundamental concept in calculus.

Step-by-Step Differentiation for sin(x)^cos(x)

  1. Set y = sin(x)^cos(x): Let y = sin(x)^cos(x). This sets the stage for logarithmic differentiation. Identifying the function that requires special treatment is essential.
  2. Take the natural logarithm of both sides: Apply the natural logarithm to both sides of the equation: ln(y) = ln(sin(x)^cos(x)). Taking the logarithm simplifies the exponentiation.
  3. Use logarithm properties to simplify: Use the property ln(a^b) = b ln(a) to rewrite the equation as ln(y) = cos(x) ln(sin(x)). This logarithmic property is crucial for simplifying expressions before differentiation.
  4. Differentiate both sides with respect to x: Differentiate both sides of the equation with respect to x. On the left side, we use implicit differentiation: (1/y) * (dy/dx). On the right side, we apply the product rule. Implicit differentiation is necessary because y is a function of x.
  5. Apply the product rule: Differentiating cos(x) ln(sin(x)) using the product rule gives us: d/dx [cos(x) ln(sin(x))] = cos'(x) ln(sin(x)) + cos(x) [ln(sin(x))]'. The product rule is essential for differentiating the product of two functions.
  6. Find the derivatives: We know that the derivative of cos(x) is -sin(x). To find the derivative of ln(sin(x)), we use the chain rule: d/dx [ln(sin(x))] = (1/sin(x)) * cos(x) = cot(x). Basic derivative rules and the chain rule are fundamental for this step.
  7. Substitute the derivatives: Substitute the derivatives back into the equation: (1/y) * (dy/dx) = -sin(x) ln(sin(x)) + cos(x) cot(x). Substituting back combines the results of the previous steps.
  8. Multiply both sides by y: Multiply both sides by y to isolate dy/dx: dy/dx = y [-sin(x) ln(sin(x)) + cos(x) cot(x)]. This step isolates the desired derivative.
  9. Substitute y = sin(x)^cos(x): Replace y with sin(x)^cos(x) to get the final derivative: dy/dx = sin(x)^cos(x) [-sin(x) ln(sin(x)) + cos(x) cot(x)]. Substituting back provides the derivative in terms of the original function.
  10. Combine the results: Combine the derivatives of both terms: dy/dx = 2x + sin(x)^cos(x) [-sin(x) ln(sin(x)) + cos(x) cot(x)]. This combines the results from differentiating each term separately.

Thus, the derivative of y = x² + sin(x)^cos(x) is dy/dx = 2x + sin(x)^cos(x) [-sin(x) ln(sin(x)) + cos(x) cot(x)]. Logarithmic differentiation is a powerful technique for handling functions where the variable appears in both the base and the exponent. This example showcases its application and utility in finding derivatives of complex functions.

Conclusion

In this comprehensive guide, we have explored how to find dy/dx for various functions and expressions. From simple algebraic functions like √x to complex expressions involving trigonometric functions and exponentiation, each example illustrates key differentiation techniques such as the power rule, chain rule, product rule, and logarithmic differentiation. Understanding these methods is crucial for mastering calculus and its applications in various fields. By following the step-by-step explanations provided, you can confidently tackle a wide range of differentiation problems and enhance your understanding of calculus.