Problem 39 - Single Variable Calculus

Question

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

Step-by-Step Solution

Answer

The derivative of the given function using logarithmic differentiation is: \[\frac{dy}{dx} = \left((\sqrt{\cos x})^{x}\right)\left[\ln{(\sqrt{\cos x})}-\frac{x\sin x}{2\sqrt{\cos x}}\right]\]

Step 1: Take the natural logarithm of both sides of the function

To start, we'll apply the natural logarithm (ln) to both sides of the equation: \[ \ln{y}=\ln{((\sqrt{\cos x})^{x})} \]

Step 2: Simplify using properties of logarithms

The given function is a composition of several functions: a square root, cosine of x, and an exponentiation. Using the properties of logarithms, we can rewrite the equation as: \[ \ln{y}=x\ln{(\sqrt{\cos x})} \]

Step 3: Apply the chain rule to implicitly differentiate

Next, we'll differentiate both sides of the equation with respect to x. The derivative of ln(y) with respect to y is (1/y), so we need to apply the chain rule and multiply this by the derivative of y with respect to x (y'): \[\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} \left(x\ln{(\sqrt{\cos x})}\right)\]

Step 4: Differentiate the right-hand side

To differentiate the right-hand side, we need to apply the product rule, which states that the derivative of u(x) * v(x) is u'(x) * v(x) + u(x) * v'(x). In our case, u(x) is x and v(x) is ln(√cos x): \[\frac{1}{y} \frac{dy}{dx} = \left(\frac{d}{dx}x\right)(\ln{(\sqrt{\cos x})})+x\left(\frac{d}{dx}\ln{(\sqrt{\cos x})}\right)\]

Step 5: Differentiate inside the logarithm

To calculate the derivative of ln(√cos x) with respect to x, we need to use the chain rule once again. We begin by differentiating the outer function, ln(f(x)): \[\frac{d}{dx}\ln{(\sqrt{\cos x})}=\frac{1}{\sqrt{\cos x}}\frac{d}{dx}\sqrt{\cos x}\] Next, we differentiate the square root, using the chain rule: \[\frac{d}{dx}\sqrt{\cos x}=\frac{1}{2}(\cos x)^{-\frac{1}{2}}\frac{d}{dx}\cos x\] Finally, we differentiate the innermost function, cos x: \[\frac{d}{dx}\cos x = -\sin x\] Now, we substitute those derivatives back into the expression for the derivative of ln(√cos x) with respect to x: \[\frac{d}{dx}\ln{(\sqrt{\cos x})} = \frac{-\sin x}{2\sqrt{\cos x}}\]

Step 6: Substitute derivatives back into the equation and simplify

Now, substitute the derivatives back into our main equation and simplify: \[\frac{1}{y} \frac{dy}{dx} = (1)(\ln{(\sqrt{\cos x})})+x\left(\frac{-\sin x}{2\sqrt{\cos x}}\right)\] Finally, multiply both sides by y to get the derivative of the function y: \[\frac{dy}{dx} = y\left[(\ln{(\sqrt{\cos x})})-\frac{x\sin x}{2\sqrt{\cos x}}\right]\] Since \(y = (\sqrt{\cos x})^{x}\), we substitute this back into our final answer: \[\frac{dy}{dx} = \left((\sqrt{\cos x})^{x}\right)\left[\ln{(\sqrt{\cos x})}-\frac{x\sin x}{2\sqrt{\cos x}}\right]\] Hence, the derivative of the given function using logarithmic differentiation is: \[\frac{dy}{dx} = \left((\sqrt{\cos x})^{x}\right)\left[\ln{(\sqrt{\cos x})}-\frac{x\sin x}{2\sqrt{\cos x}}\right]\]

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus used to differentiate composite functions. These are functions within other functions, like \((f(g(x)))\). The rule helps us determine the derivative of such functions in a systematic way.
To apply the Chain Rule:

• Differentiating the outer function first, leaving the inner function unchanged.
• Multiply the result by the derivative of the inner function.

In our exercise, we encountered composite functions such as \(\ln(\sqrt{\cos x})\). First, we differentiated \(\ln(u)\), where \(u=\sqrt{\cos x}\), and then multiplied by the derivative of \(u\) with respect to \(x\). This divides the differentiation problem into easier steps.
If understanding derivatives of complex expressions seems daunting, remember that the Chain Rule is your best friend, simplifying the process through step-by-step peeling of layers in a function.

Product Rule

The Product Rule comes into play when differentiating functions that are products of two or more functions. Say, we have a function \(h(x) = u(x) \, v(x)\), the Product Rule states:

• The derivative, \(h'(x)\), is \(u'(x)v(x) + u(x)v'(x)\).

In our example, we used the Product Rule to handle \(x\ln(\sqrt{\cos x})\). Here, \(u(x) = x\) and \(v(x) = \ln(\sqrt{\cos x})\).
Applying the rule:

• The derivative of \(x\) is \(1\).
• We already calculated the derivative of \(\ln(\sqrt{\cos x})\) using the Chain Rule.

Finally, assembling these pieces using the Product Rule, we arrived at the differentiated expression. Learning to spot products within functions is key, allowing you to strategically apply the Product Rule in calculus problems.

Implicit Differentiation

Implicit Differentiation is a technique used when functions are not given in the form \(y=f(x)\) directly. Instead, they are implied within equations. We use this method to find the derivative of \(y\) with respect to \(x\) without explicitly solving for \(y\).
In our problem, we started by taking the natural logarithm of both sides, transforming the expression into a form suitable for implicit differentiation. Then, we implicitly differentiated with respect to \(x\), treating \(y\) as a function of \(x\).

• When differentiating \(\ln{y}\), we expressed it as \(\frac{1}{y} \frac{dy}{dx}\) using the Chain Rule.

Implicit differentiation proves invaluable when dealing with complex equations, enabling a methodical differentiation step when \(y\) cannot be isolated easily.

Derivative of Exponential Functions

The derivative of exponential functions is an essential topic and arises frequently in calculus. Exponential functions have the form \((a^x)\), with \(a\) as a constant. The core idea is that the derivative of \(a^x\) is \(a^x\ln(a)\).
In our problem, the function is \(y=(\sqrt{\cos x})^x\). Unlike simple exponential functions, complexity is introduced with both the base and the exponent containing the variable \(x\). This complexity is why we employed logarithmic differentiation.
When differentiating exponential functions:

• Understand how the base and the exponent contribute to the expression.
• Use logarithmic differentiation to simplify especially if both involve \(x\).

This ensures you can derive even the trickiest of exponential functions by systematically breaking down each component.