Problem 76 - Calculus for Biology and Medicine

Question

Use logarithmic differentiation to find the first derivative of the given functions. $$ y=(\cos x)^{x} $$

Step-by-Step Solution

Answer

The derivative of the function is \( \frac{dy}{dx} = (\cos x)^x \left( \ln(\cos x) - x \tan x \right) \).

Step 1: Apply the Natural Logarithm

Take the natural logarithm of both sides of the equation to simplify the expression:\[ \ln y = \ln ((\cos x)^x) \]Using the properties of logarithms, this becomes:\[ \ln y = x \ln(\cos x) \].

Step 2: Differentiate Implicitly

Differentiate both sides with respect to \(x\). Remember to use the product rule on the right-hand side:\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln(\cos x)) \]The left side becomes:\[ \frac{1}{y} \frac{dy}{dx} \].The right side using the product rule is:\[ 1 \cdot \ln(\cos x) + x \cdot \left(-\frac{\sin x}{\cos x}\right) \]Simplify to:\[ \ln(\cos x) - x \tan x \].

Step 3: Solve for the Derivative

Equating the derivatives from the implicit differentiation step gives:\[ \frac{1}{y} \frac{dy}{dx} = \ln(\cos x) - x \tan x \].Multiply through by \(y\) to solve for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = y \cdot \left( \ln(\cos x) - x \tan x \right) \].

Step 4: Substitute Back for y

Recall that \(y = (\cos x)^x\), so substitute this back into the derivative:\[ \frac{dy}{dx} = (\cos x)^x \left( \ln(\cos x) - x \tan x \right) \].

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when differentiating equations that are not solved explicitly for one variable in terms of another. In our case, we initially take the natural logarithm of the function to simplify the differentiation process.

Here's the step-by-step breakdown of implicit differentiation as applied in our exercise:

• First, you take the natural logarithm of both sides, making the exponent on the function easier to handle. This is because logarithms allow us to bring the exponent down as a coefficient.
• Once this transformation is done, we differentiate both sides of the equation. The key here is to remember that on the left side, we have \(\ln y\), which differentiates to \[\frac{1}{y} \frac{dy}{dx}\], following the chain rule.
• On the right side, you are dealing with the derivative of products and sums, which brings us to the product rule next.

By understanding and using implicit differentiation, calculations involving more complex functions become more manageable. This concept is especially helpful when the function isn’t in the typical y = f(x) form, as seen in logarithmic differentiation scenarios.

Product Rule

The product rule is crucial when dealing with derivatives of expressions that are products of two functions. In the given exercise, once we applied the logarithm, the right-hand side became the product \(x \ln(\cos x)\).

To differentiate this product, the product rule states that if you have two functions u(x) and v(x), their derivative is:

• \[ \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \]

In our example:

• First function (u) = x, whose derivative u' = 1
• Second function (v) = \(\ln(\cos x)\), and its derivative v' involves the chain rule and differentiates to \(-\tan x\).
• Using the rule gives \(1 \cdot \ln(\cos x) + x \cdot (-\tan x)\) for the complete derivative of the product.

The product rule makes it feasible to tackle composite expressions by breaking them down into more straightforward parts, which is just what we did here. It's a reliable tool in calculus whenever you face products.

Trigonometric Functions

Trigonometric functions are a fundamental part of calculus, and they often require additional rules when taking derivatives. In the context of our problem, the trigonometric function involved is \(\cos x\).

The derivative of trigonometric functions requires us to be familiar with a few essential facts:

• \(\frac{d}{dx}(\cos x) = -\sin x\)
• \(\tan x\), which is \(\frac{\sin x}{\cos x}\), is also a part of the differentiation as seen in the exercise.

When working within the logarithmic form, we used the chain rule and some identities for calculations:

• \(\ln(\cos x)\)'s derivative results in \(-\tan x\), which combines the basic derivatives of trigonometric functions and knowledge of the chain rule.

Mastering derivatives of trigonometric functions, through exercises that combine them with logarithms and other calculus rules, build a solid foundation for solving more complex calculations in calculus.