Quotient Rule Differentiation Of Y = Ln(x) / Sec(x)

by ADMIN 52 views

The quotient rule is a fundamental concept in calculus that provides a method for differentiating functions that are expressed as the ratio of two other functions. This article delves into the application of the quotient rule, specifically focusing on the differentiation of the function $y = \frac{\ln x}{\sec x}$. Mastering the quotient rule is crucial for anyone studying calculus, as it appears frequently in various mathematical problems. We will break down the process step by step, ensuring a clear understanding of each stage involved.

The quotient rule is mathematically stated as follows: If we have a function $y$ that is defined as $y = \frac{u(x)}{v(x)}$, where $u(x)$ and $v(x)$ are both differentiable functions, then the derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$, is given by:

dydx=v(x)β‹…uβ€²(x)βˆ’u(x)β‹…vβ€²(x)[v(x)]2\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}

In simpler terms, the derivative of a quotient is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. This formula is essential for tackling functions that are expressed as fractions where both the numerator and the denominator are functions of $x$.

Applying the Quotient Rule to y = ln(x) / sec(x)

To differentiate $y = \frac{\ln x}{\sec x}$ using the quotient rule, we first identify the numerator and denominator functions. Let's define:

  • u(x) = \ln x$ (the numerator)

  • v(x) = \sec x$ (the denominator)

Next, we need to find the derivatives of both $u(x)$ and $v(x)$ with respect to $x$.

Finding u'(x)

The derivative of $u(x) = \ln x$ is a standard derivative that you should be familiar with. The derivative of the natural logarithm function, $\ln x$, is simply $\frac{1}{x}$. Therefore,

uβ€²(x)=ddx(ln⁑x)=1xu'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x}

Finding v'(x)

The derivative of $v(x) = \sec x$ is another standard derivative. Recall that $\sec x$ is the reciprocal of $\cos x$, i.e., $\sec x = \frac{1}{\cos x}$. The derivative of $\sec x$ is $\sec x \tan x$. Thus,

vβ€²(x)=ddx(sec⁑x)=sec⁑xtan⁑xv'(x) = \frac{d}{dx}(\sec x) = \sec x \tan x

Now that we have $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$, we can apply the quotient rule formula:

dydx=v(x)β‹…uβ€²(x)βˆ’u(x)β‹…vβ€²(x)[v(x)]2\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}

Substitute the functions and their derivatives into the formula:

dydx=sec⁑xβ‹…1xβˆ’ln⁑xβ‹…sec⁑xtan⁑x(sec⁑x)2\frac{dy}{dx} = \frac{\sec x \cdot \frac{1}{x} - \ln x \cdot \sec x \tan x}{(\sec x)^2}

Simplifying the Derivative

The expression we obtained in the previous step can be simplified to a more manageable form. We have:

dydx=sec⁑xβ‹…1xβˆ’ln⁑xβ‹…sec⁑xtan⁑xsec⁑2x\frac{dy}{dx} = \frac{\sec x \cdot \frac{1}{x} - \ln x \cdot \sec x \tan x}{\sec^2 x}

To simplify this, we can factor out $\sec x$ from the numerator:

dydx=sec⁑x(1xβˆ’ln⁑xtan⁑x)sec⁑2x\frac{dy}{dx} = \frac{\sec x \left(\frac{1}{x} - \ln x \tan x\right)}{\sec^2 x}

Now, we can cancel out one factor of $\sec x$ from the numerator and the denominator:

dydx=1xβˆ’ln⁑xtan⁑xsec⁑x\frac{dy}{dx} = \frac{\frac{1}{x} - \ln x \tan x}{\sec x}

To further simplify, we can multiply both the numerator and the denominator by $x$ to eliminate the fraction within the fraction:

dydx=1βˆ’xln⁑xtan⁑xxsec⁑x\frac{dy}{dx} = \frac{1 - x \ln x \tan x}{x \sec x}

This is the simplified form of the derivative of $y = \frac{\ln x}{\sec x}$ with respect to $x$.

Alternative Simplification Steps

Alternatively, instead of factoring out $\sec x$ in the early simplification stage, we could manipulate the expression by multiplying both the numerator and the denominator by $\cos^2 x$. Recall that $\sec x = \frac{1}{\cos x}$, so $\sec^2 x = \frac{1}{\cos^2 x}$. Multiplying by $\cos^2 x$ helps to eliminate the secant terms and express the derivative in terms of sines and cosines.

Starting from:

dydx=sec⁑xβ‹…1xβˆ’ln⁑xβ‹…sec⁑xtan⁑xsec⁑2x\frac{dy}{dx} = \frac{\sec x \cdot \frac{1}{x} - \ln x \cdot \sec x \tan x}{\sec^2 x}

Multiply both the numerator and the denominator by $\cos^2 x$:

dydx=cos⁑2x(sec⁑xβ‹…1xβˆ’ln⁑xβ‹…sec⁑xtan⁑x)cos⁑2xsec⁑2x\frac{dy}{dx} = \frac{\cos^2 x \left(\sec x \cdot \frac{1}{x} - \ln x \cdot \sec x \tan x\right)}{\cos^2 x \sec^2 x}

Since $\cos^2 x \sec^2 x = 1$, the denominator simplifies to 1. Distribute $\cos^2 x$ in the numerator:

dydx=cos⁑2xsec⁑xβ‹…1xβˆ’cos⁑2xln⁑xsec⁑xtan⁑x\frac{dy}{dx} = \cos^2 x \sec x \cdot \frac{1}{x} - \cos^2 x \ln x \sec x \tan x

Now, use the identities $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$:

dydx=cos⁑xβ‹…1xβˆ’cos⁑2xln⁑xβ‹…1cos⁑xβ‹…sin⁑xcos⁑x\frac{dy}{dx} = \cos x \cdot \frac{1}{x} - \cos^2 x \ln x \cdot \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}

Simplify:

dydx=cos⁑xxβˆ’ln⁑xsin⁑x\frac{dy}{dx} = \frac{\cos x}{x} - \ln x \sin x

To combine these terms into a single fraction, multiply the second term by $\frac{x}{x}$:

dydx=cos⁑xβˆ’xln⁑xsin⁑xx\frac{dy}{dx} = \frac{\cos x - x \ln x \sin x}{x}

This is another simplified form of the derivative. While it looks different from the previous simplified form, both are mathematically equivalent. The choice of which form to use often depends on the context of the problem or the desired form for further calculations.

Common Mistakes to Avoid

When applying the quotient rule, there are several common mistakes that students often make. Being aware of these can help you avoid errors and ensure accurate differentiation.

Incorrect Application of the Quotient Rule Formula

The most common mistake is misremembering or misapplying the quotient rule formula. The correct formula is:

dydx=v(x)β‹…uβ€²(x)βˆ’u(x)β‹…vβ€²(x)[v(x)]2\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}

A frequent error is reversing the terms in the numerator or forgetting to square the denominator. Always double-check the formula before applying it to ensure you have the correct order and operations.

Incorrectly Differentiating u(x) or v(x)

Another common mistake is incorrectly finding the derivatives of $u(x)$ and $v(x)$. For example, when differentiating $\ln x$, some students might incorrectly write the derivative as $x$ or $-\frac{1}{x}$. Similarly, the derivative of $\sec x$ is often confused with other trigonometric derivatives. Ensure you know the derivatives of common functions by heart or have a reference sheet handy.

Algebraic Errors in Simplification

After applying the quotient rule, the resulting expression often requires simplification. Algebraic errors, such as incorrect factoring or canceling terms, are common at this stage. Always take your time when simplifying and double-check each step. Factoring out common terms, like we did with $\sec x$ in our example, can often make the expression easier to manage.

Forgetting the Chain Rule

If $u(x)$ or $v(x)$ are composite functions (functions within functions), you'll need to apply the chain rule in conjunction with the quotient rule. For instance, if we had $y = \frac{\ln(x^2)}{\sec x}$, the derivative of $\ln(x^2)$ would require the chain rule. Forgetting this can lead to an incorrect derivative. Always identify composite functions and apply the chain rule where necessary.

Not Simplifying the Final Answer

While obtaining the correct derivative is crucial, simplifying the final answer is equally important. A simplified answer is easier to interpret and use in further calculations. Common simplifications include factoring, canceling terms, and using trigonometric identities. In our example, we simplified the derivative by factoring out $\sec x$ and by multiplying by $\cos^2 x$. Always aim to present your answer in the simplest form possible.

Practice Problems

To solidify your understanding of the quotient rule, working through practice problems is essential. Here are a few problems you can try:

  1. Differentiate $y = \frac{x^2}{\cos x}$
  2. Differentiate $y = \frac{ex}{x3}$
  3. Differentiate $y = \frac{\sin x}{x}$
  4. Differentiate $y = \frac{x}{\ln x}$
  5. Differentiate $y = \frac{\tan x}{x^2 + 1}$

Work through these problems step by step, paying close attention to each stage of the quotient rule application and simplification. Check your answers against known solutions or use online calculators to verify your results.

Conclusion

The quotient rule is a vital tool in calculus for differentiating functions expressed as the ratio of two other functions. By understanding and applying the quotient rule correctly, you can solve a wide range of differentiation problems. In this article, we walked through the differentiation of $y = \frac{\ln x}{\sec x}$ using the quotient rule, highlighting each step and providing simplifications. We also discussed common mistakes to avoid and offered practice problems to enhance your skills. Mastering the quotient rule will significantly improve your calculus proficiency and problem-solving abilities.

Remember, practice is key to mastering calculus concepts. Keep working on various problems, and you'll become more confident and proficient in applying the quotient rule and other differentiation techniques. Happy differentiating!