Differentiating Y = 3log3x - Cot2x + 3/e^(-x) + 7/x^2 - 2lnx + (sin2x)/cosx

In this comprehensive guide, we will delve into the process of differentiating a complex function with respect to $x$. Our focus will be on the function:

$y=3 \log _3 x-\cot 2 x+\frac{3}{e^{-x}}+\frac{7}{x^2}-2 \ln x+\frac{\sin 2 x}{\cos x}$

We will break down the differentiation process step-by-step, applying relevant calculus rules and techniques to arrive at the final derivative. This guide is designed to be accessible to students and anyone interested in enhancing their understanding of calculus.

Understanding the Function

Before we begin differentiating, let's take a closer look at the function itself. It comprises several terms, each involving different mathematical functions:

1. $3 \log _3 x$: This term involves a logarithmic function with base 3.
2. $\cot 2 x$: This term involves the cotangent function, which is the reciprocal of the tangent function, with an argument of $2x$.
3. $\frac{3}{e^{-x}}$: This term involves an exponential function in the denominator, which can be simplified.
4. $\frac{7}{x^2}$: This term involves a power function in the denominator.
5. $-2 \ln x$: This term involves the natural logarithm function.
6. $\frac{\sin 2 x}{\cos x}$: This term involves trigonometric functions, which can be simplified using trigonometric identities.

To effectively differentiate this function, we will need to apply various differentiation rules, including the chain rule, quotient rule, and the derivatives of logarithmic, trigonometric, and exponential functions.

Step 1 Differentiate $3 \log _3 x$

Our first step in differentiating the complex function is to address the logarithmic term $3 \log _3 x$. To effectively differentiate this term, we will employ the change of base formula for logarithms and the standard differentiation rule for natural logarithms. The change of base formula allows us to convert the logarithm from base 3 to the natural logarithm (base $e$), which is more convenient for differentiation. This conversion is crucial for simplifying the differentiation process and applying the well-known derivative of the natural logarithm.

The change of base formula is given by:

$\log _a b = \frac{\log _c b}{\log _c a}$

Applying this formula to our term, we get:

$3 \log _3 x = 3 \frac{\ln x}{\ln 3}$

Now, we can differentiate this expression with respect to $x$. Recall that the derivative of $\ln x$ with respect to $x$ is $\frac{1}{x}$. Also, note that $\ln 3$ is a constant. Therefore, the derivative of $3 \frac{\ln x}{\ln 3}$ with respect to $x$ is:

$\frac{d}{dx} \left( 3 \frac{\ln x}{\ln 3} \right) = \frac{3}{\ln 3} \frac{d}{dx} (\ln x) = \frac{3}{\ln 3} \cdot \frac{1}{x} = \frac{3}{x \ln 3}$

Thus, the derivative of the first term, $3 \log _3 x$, is $\frac{3}{x \ln 3}$. This result is a crucial building block for finding the overall derivative of the given complex function. By understanding and correctly applying the change of base formula and the derivative of the natural logarithm, we have successfully differentiated the first term of our function. This methodical approach is key to tackling more complex derivatives in the subsequent steps.

Step 2 Differentiate $-\cot 2 x$

Next, we will differentiate the term $-\cot 2x$. This involves differentiating a trigonometric function, specifically the cotangent function. To do this effectively, we will use the chain rule, a fundamental concept in calculus that allows us to differentiate composite functions. The chain rule is particularly useful when dealing with trigonometric functions that have arguments other than a simple $x$, as is the case here with $2x$.

Recall that the derivative of $\cot x$ with respect to $x$ is $-\csc^2 x$. Applying the chain rule, we need to consider the derivative of the argument $2x$ as well. The chain rule states that if we have a function $y = f(g(x))$, then its derivative with respect to $x$ is given by:

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

In our case, $f(u) = -\cot u$ and $g(x) = 2x$. Thus, we first find the derivative of $f(u)$ with respect to $u$, which is:

$\frac{d}{du} (-\cot u) = -(-\csc^2 u) = \csc^2 u$

Next, we find the derivative of $g(x)$ with respect to $x$:

$\frac{d}{dx} (2x) = 2$

Now, applying the chain rule, we multiply these two results together, substituting $2x$ back in for $u$:

$\frac{d}{dx} (-\cot 2x) = \csc^2 (2x) \cdot 2 = 2 \csc^2 2x$

Therefore, the derivative of $-\cot 2x$ with respect to $x$ is $2 \csc^2 2x$. This step demonstrates the application of the chain rule in differentiating trigonometric functions, a crucial skill in calculus. By correctly identifying the inner and outer functions and applying the chain rule, we have successfully found the derivative of the second term in our complex function.

Step 3 Differentiate $\frac{3}{e^{-x}}$

Moving on, we will now differentiate the term $\frac{3}{e^{-x}}$. This term involves an exponential function in the denominator. To simplify the differentiation process, we can first rewrite the term using properties of exponents. This preliminary step will make the subsequent application of differentiation rules more straightforward and less prone to errors.

Recall that $\frac{1}{e^{-x}}$ is equivalent to $e^x$. Therefore, we can rewrite the term as:

$\frac{3}{e^{-x}} = 3e^x$

Now, differentiating this simplified expression with respect to $x$ is much easier. We know that the derivative of $e^x$ with respect to $x$ is simply $e^x$. Thus, the derivative of $3e^x$ with respect to $x$ is:

$\frac{d}{dx} (3e^x) = 3 \frac{d}{dx} (e^x) = 3e^x$

Therefore, the derivative of $\frac{3}{e^{-x}}$ with respect to $x$ is $3e^x$. This step highlights the importance of simplifying expressions before differentiating, especially when dealing with exponential functions. By rewriting the term using exponent rules, we were able to easily apply the standard derivative of the exponential function and arrive at the correct result. This technique is valuable in calculus for handling various types of functions and making the differentiation process more manageable.

Step 4 Differentiate $\frac{7}{x^2}$

Now, let's differentiate the term $\frac{7}{x^2}$. This term involves a power function in the denominator. Similar to the previous step, we can simplify the differentiation process by rewriting the term using properties of exponents. This will allow us to apply the power rule of differentiation more easily.

We can rewrite $\frac{7}{x^2}$ as $7x^{-2}$. Now, we can apply the power rule, which states that if $y = ax^n$, then $\frac{dy}{dx} = nax^{n-1}$. In our case, $a = 7$ and $n = -2$. Applying the power rule, we get:

$\frac{d}{dx} (7x^{-2}) = 7 \frac{d}{dx} (x^{-2}) = 7(-2)x^{-2-1} = -14x^{-3}$

We can rewrite $x^{-3}$ as $\frac{1}{x^3}$, so the derivative becomes:

$-14x^{-3} = -\frac{14}{x^3}$

Therefore, the derivative of $\frac{7}{x^2}$ with respect to $x$ is $-\frac{14}{x^3}$. This step demonstrates the application of the power rule in differentiating functions with negative exponents. By rewriting the term with a negative exponent, we were able to directly apply the power rule and find the derivative. This technique is a fundamental tool in calculus for differentiating polynomial and rational functions.

Step 5 Differentiate $-2 \ln x$

Next, we will differentiate the term $-2 \ln x$. This term involves the natural logarithm function. The derivative of the natural logarithm function is a fundamental concept in calculus, and we will apply it directly here.

Recall that the derivative of $\ln x$ with respect to $x$ is $\frac{1}{x}$. Therefore, the derivative of $-2 \ln x$ with respect to $x$ is:

$\frac{d}{dx} (-2 \ln x) = -2 \frac{d}{dx} (\ln x) = -2 \cdot \frac{1}{x} = -\frac{2}{x}$

Thus, the derivative of $-2 \ln x$ with respect to $x$ is $-\frac{2}{x}$. This step is a straightforward application of the derivative of the natural logarithm function. It reinforces the importance of knowing the basic derivatives of common functions, such as the natural logarithm, for tackling more complex differentiation problems. This result is a key component in finding the overall derivative of our complex function.

Step 6 Differentiate $\frac{\sin 2 x}{\cos x}$

Finally, we will differentiate the term $\frac{\sin 2 x}{\cos x}$. This term involves trigonometric functions and requires careful simplification before differentiation. We can use trigonometric identities to simplify the expression, making it easier to differentiate using the quotient rule or other appropriate methods.

Recall the double angle identity for sine:

$\sin 2x = 2 \sin x \cos x$

Using this identity, we can rewrite the term as:

$\frac{\sin 2 x}{\cos x} = \frac{2 \sin x \cos x}{\cos x}$

Now, we can cancel out the $\cos x$ terms, provided that $\cos x \neq 0$:

$\frac{2 \sin x \cos x}{\cos x} = 2 \sin x$

Now, differentiating $2 \sin x$ with respect to $x$ is straightforward. The derivative of $\sin x$ with respect to $x$ is $\cos x$. Therefore, the derivative of $2 \sin x$ with respect to $x$ is:

$\frac{d}{dx} (2 \sin x) = 2 \frac{d}{dx} (\sin x) = 2 \cos x$

Therefore, the derivative of $\frac{\sin 2 x}{\cos x}$ with respect to $x$ is $2 \cos x$. This step demonstrates the power of using trigonometric identities to simplify expressions before differentiating. By applying the double angle identity for sine, we were able to simplify the term significantly and easily find its derivative. This technique is crucial for efficiently differentiating trigonometric functions and avoiding the complexities of the quotient rule in this case.

Final Result

Now, we combine the derivatives of each term to find the derivative of the entire function:

$\frac{d}{dx} \left( 3 \log _3 x - \cot 2 x + \frac{3}{e^{-x}} + \frac{7}{x^2} - 2 \ln x + \frac{\sin 2 x}{\cos x} \right) = \frac{3}{x \ln 3} + 2 \csc^2 2x + 3e^x - \frac{14}{x^3} - \frac{2}{x} + 2 \cos x$

Therefore, the derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = \frac{3}{x \ln 3} + 2 \csc^2 2x + 3e^x - \frac{14}{x^3} - \frac{2}{x} + 2 \cos x$

This comprehensive guide has walked you through the process of differentiating a complex function step-by-step. By breaking down the function into individual terms and applying the appropriate differentiation rules and techniques, we have successfully found the derivative. This methodical approach is essential for tackling complex calculus problems and building a strong understanding of differentiation.