Find Dy/dx For Y = X^(tan X) + (sin X)^(cos X)

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This article will walk you through the steps to find the derivative, denoted as dy/dx, of the function y = x^(tan x) + (sin x)^(cos x). This problem involves differentiating a function where both the base and the exponent are functions of x, requiring us to use logarithmic differentiation. We'll break down the problem into smaller, manageable parts and then combine the results.

Breaking Down the Problem

The given function is a sum of two functions: x^(tan x) and (sin x)^(cos x). Let's denote these as u and v, respectively, so that:

y = u + v

Where:

  • u = x^(tan x)
  • v = (sin x)^(cos x)

To find dy/dx, we will first find du/dx and dv/dx separately and then add them together, using the basic rule of calculus that the derivative of a sum is the sum of the derivatives:

dy/dx = du/dx + dv/dx

This approach simplifies the problem, allowing us to focus on each term individually. The core technique we'll employ is logarithmic differentiation, a powerful method when dealing with functions of the form f(x)^g(x), where both f(x) and g(x) are functions of x. This method involves taking the natural logarithm of both sides of the equation, which transforms the exponentiation into a more manageable multiplication. This will be particularly useful as we tackle the complexities of x^(tan x) and (sin x)^(cos x).

Differentiating u = x^(tan x)

To find du/dx where u = x^(tan x), we'll employ logarithmic differentiation. This technique is indispensable when dealing with functions where both the base and the exponent are functions of x, as is the case here. By taking the natural logarithm of both sides, we transform the exponential function into a product, which is far easier to differentiate.

  1. Take the natural logarithm of both sides:

    ln(u) = ln(x^(tan x))

    Using the logarithm power rule, we can rewrite the right side:

    ln(u) = tan(x) * ln(x)

  2. Differentiate both sides with respect to x:

    We'll use the chain rule on the left side and the product rule on the right side. The derivative of ln(u) with respect to x is (1/u) * (du/dx). The product rule states that the derivative of (f(x) * g(x)) is f'(x) * g(x) + f(x) * g'(x). In our case, f(x) = tan(x) and g(x) = ln(x).

    (1/u) * (du/dx) = (d/dx [tan(x)]) * ln(x) + tan(x) * (d/dx [ln(x)])

    Recall that the derivative of tan(x) is sec^2(x) and the derivative of ln(x) is 1/x. Substituting these, we get:

    (1/u) * (du/dx) = sec^2(x) * ln(x) + tan(x) * (1/x)

  3. Solve for du/dx:

    Multiply both sides by u to isolate du/dx:

    du/dx = u * [sec^2(x) * ln(x) + tan(x) / x]

    Now, substitute u = x^(tan x) back into the equation:

    du/dx = x^(tan x) * [sec^2(x) * ln(x) + tan(x) / x]

This gives us the derivative of the first part of our function. We've successfully navigated the complexities of differentiating x^(tan x) by strategically applying logarithmic differentiation and the product rule. This meticulous step-by-step process illustrates the power and elegance of calculus in tackling seemingly intricate problems. The next phase involves applying a similar approach to differentiate the second term, (sin x)^(cos x), further showcasing the versatility of these techniques.

Differentiating v = (sin x)^(cos x)

Now, let's tackle the second part of our problem: finding dv/dx where v = (sin x)^(cos x). Just as with the previous term, we will again rely on logarithmic differentiation. This method proves invaluable when dealing with functions where both the base and the exponent are functions of x, transforming a complex exponential problem into a more manageable algebraic one.

  1. Take the natural logarithm of both sides:

    ln(v) = ln((sin x)^(cos x))

    Using the power rule of logarithms, we bring the exponent down as a coefficient:

    ln(v) = cos(x) * ln(sin x)

  2. Differentiate both sides with respect to x:

    We apply the chain rule on the left side, noting that the derivative of ln(v) with respect to x is (1/v) * (dv/dx). On the right side, we encounter another product, requiring the product rule. This rule states that the derivative of (f(x) * g(x)) is f'(x) * g(x) + f(x) * g'(x). Here, f(x) = cos(x) and g(x) = ln(sin x).

    (1/v) * (dv/dx) = (d/dx [cos(x)]) * ln(sin x) + cos(x) * (d/dx [ln(sin x)])

    The derivative of cos(x) is -sin(x). To find the derivative of ln(sin x), we need to use the chain rule again. The derivative of ln(u) is 1/u, so the derivative of ln(sin x) is (1/sin x) * (d/dx [sin x]) = (1/sin x) * cos(x) = cot(x). Substituting these derivatives into our equation:

    (1/v) * (dv/dx) = -sin(x) * ln(sin x) + cos(x) * cot(x)

  3. Solve for dv/dx:

    Multiply both sides by v to isolate dv/dx:

    dv/dx = v * [-sin(x) * ln(sin x) + cos(x) * cot(x)]

    Substitute v = (sin x)^(cos x) back into the equation:

    dv/dx = (sin x)^(cos x) * [-sin(x) * ln(sin x) + cos(x) * cot(x)]

We've now successfully determined the derivative of the second part of our function, (sin x)^(cos x). This process reinforces the importance of logarithmic differentiation when dealing with variable exponents and showcases the necessity of a methodical, step-by-step approach to complex calculus problems. The skillful application of the product rule and chain rule, combined with logarithmic differentiation, has allowed us to dissect and conquer this seemingly daunting derivative. With both du/dx and dv/dx in hand, we are now poised to combine these results to find the complete solution: dy/dx.

Combining the Results to Find dy/dx

Having calculated du/dx and dv/dx, we are now ready to find dy/dx. Recall that we initially decomposed the original function y into two parts, u and v, where y = u + v. We established that:

dy/dx = du/dx + dv/dx

This elegant expression allows us to simply add the two derivatives we've painstakingly computed, bringing us to the final solution.

We previously found:

  • du/dx = x^(tan x) * [sec^2(x) * ln(x) + tan(x) / x]
  • dv/dx = (sin x)^(cos x) * [-sin(x) * ln(sin x) + cos(x) * cot(x)]

Now, we substitute these expressions into our equation for dy/dx:

dy/dx = x^(tan x) * [sec^2(x) * ln(x) + tan(x) / x] + (sin x)^(cos x) * [-sin(x) * ln(sin x) + cos(x) * cot(x)]

This is the final derivative of the function y = x^(tan x) + (sin x)^(cos x). This expression, while seemingly complex, is a testament to the power of calculus in handling intricate functions. It represents the instantaneous rate of change of y with respect to x, encapsulating the dynamic relationship between these variables. The solution underscores the effectiveness of breaking down a complex problem into smaller, manageable parts, applying appropriate differentiation techniques, and then systematically recombining the results.

Conclusion

In conclusion, we successfully found dy/dx for the function y = x^(tan x) + (sin x)^(cos x) by employing logarithmic differentiation, the product rule, and the chain rule. The final answer is:

dy/dx = x^(tan x) * [sec^2(x) * ln(x) + tan(x) / x] + (sin x)^(cos x) * [-sin(x) * ln(sin x) + cos(x) * cot(x)]

This problem highlights the importance of recognizing function structure and strategically applying differentiation techniques. Logarithmic differentiation is particularly powerful when dealing with functions where both the base and exponent are functions of x. Remember to always double-check your work and ensure you've applied the rules of calculus correctly. For further exploration of calculus concepts and techniques, you can visit reputable resources such as Khan Academy's Calculus section.