Compute The Derivative Of The Function:$\[ G(z) = \frac{5}{\tan(z) - \cot(z)} \\]Express Numbers In Exact Form.$\[ G^{\prime}(z) = \\]

Derivative of a Trigonometric Function: Computing G'(z)

In this article, we will delve into the world of calculus and explore the concept of finding the derivative of a given function. Specifically, we will focus on computing the derivative of the function $G(z) = \frac{5}{\tan(z) - \cot(z)}$. This function involves trigonometric functions, which are essential in mathematics and have numerous applications in various fields.

Before we proceed with finding the derivative, let's break down the given function and understand its components. The function $G(z)$ is defined as:

$$G(z) = \frac{5}{\tan(z) - \cot(z)}$$

Here, $\tan(z)$ and $\cot(z)$ are trigonometric functions, where $\tan(z) = \frac{\sin(z)}{\cos(z)}$ and $\cot(z) = \frac{\cos(z)}{\sin(z)}$. The function $G(z)$ can be rewritten as:

$$G(z) = \frac{5}{\frac{\sin(z)}{\cos(z)} - \frac{\cos(z)}{\sin(z)}}$$

To simplify the function, we can combine the fractions in the denominator:

$$G(z) = \frac{5}{\frac{\sin^2(z) - \cos^2(z)}{\sin(z)\cos(z)}}$$

Using the identity $\sin^2(z) - \cos^2(z) = -\cos(2z)$, we can rewrite the function as:

$$G(z) = \frac{5}{\frac{-\cos(2z)}{\sin(z)\cos(z)}}$$

Simplifying further, we get:

$$G(z) = \frac{5\sin(z)\cos(z)}{-\cos(2z)}$$

Now that we have simplified the function, we can proceed with finding its derivative. To do this, we will apply the quotient rule of differentiation, which states that if $f(z) = \frac{g(z)}{h(z)}$, then $f'(z) = \frac{h(z)g'(z) - g(z)h'(z)}{[h(z)]^2}$.

Applying the quotient rule to our function, we get:

$$G'(z) = \frac{(-\cos(2z))\frac{d}{dz}\left(5\sin(z)\cos(z)\right) - 5\sin(z)\cos(z)\frac{d}{dz}\left(-\cos(2z)\right)}{(-\cos(2z))^2}$$

To find the derivative of the numerator, we will apply the product rule of differentiation, which states that if $f(z) = g(z)h(z)$, then $f'(z) = g'(z)h(z) + g(z)h'(z)$.

Applying the product rule to the numerator, we get:

$$\frac{d}{dz}\left(5\sin(z)\cos(z)\right) = 5\cos^2(z) - 5\sin^2(z)$$

To find the derivative of the denominator, we will apply the chain rule of differentiation, which states that if $f(z) = g(h(z))$, then $f'(z) = g'(h(z))h'(z)$.

Applying the chain rule to the denominator, we get:

$$\frac{d}{dz}\left(-\cos(2z)\right) = 2\sin(2z)$$

Now that we have found the derivatives of the numerator and denominator, we can substitute them back into the expression for $G'(z)$:

$$G'(z) = \frac{(-\cos(2z))(5\cos^2(z) - 5\sin^2(z)) - 5\sin(z)\cos(z)(2\sin(2z))}{(-\cos(2z))^2}$$

To simplify the derivative, we can expand the numerator and denominator:

$$G'(z) = \frac{-5\cos^3(2z) + 5\sin^2(2z)\cos(2z) - 10\sin^2(z)\cos^2(z) - 10\sin^2(z)\cos^2(z)}{\cos^2(2z)}$$

Combining like terms, we get:

$$G'(z) = \frac{-5\cos^3(2z) + 5\sin^2(2z)\cos(2z) - 20\sin^2(z)\cos^2(z)}{\cos^2(2z)}$$

The derivative of the function $G(z) = \frac{5}{\tan(z) - \cot(z)}$ is:

$$G'(z) = \frac{-5\cos^3(2z) + 5\sin^2(2z)\cos(2z) - 20\sin^2(z)\cos^2(z)}{\cos^2(2z)}$$

In our previous article, we explored the concept of finding the derivative of a given function, specifically the function $G(z) = \frac{5}{\tan(z) - \cot(z)}$. We broke down the function, simplified it, and then found its derivative using the quotient rule, product rule, and chain rule of differentiation. In this article, we will address some common questions and concerns related to the derivative of the function $G(z)$.

Q: What is the derivative of the function $G(z)$?

A: The derivative of the function $G(z) = \frac{5}{\tan(z) - \cot(z)}$ is:

$$G'(z) = \frac{-5\cos^3(2z) + 5\sin^2(2z)\cos(2z) - 20\sin^2(z)\cos^2(z)}{\cos^2(2z)}$$

Q: How did you simplify the function $G(z)$?

A: To simplify the function $G(z)$, we combined the fractions in the denominator and used the identity $\sin^2(z) - \cos^2(z) = -\cos(2z)$ to rewrite the function as:

$$G(z) = \frac{5\sin(z)\cos(z)}{-\cos(2z)}$$

Q: What is the quotient rule of differentiation?

A: The quotient rule of differentiation states that if $f(z) = \frac{g(z)}{h(z)}$, then $f'(z) = \frac{h(z)g'(z) - g(z)h'(z)}{[h(z)]^2}$.

Q: How did you apply the quotient rule to the function $G(z)$?

A: To apply the quotient rule to the function $G(z)$, we identified the numerator and denominator as $g(z) = 5\sin(z)\cos(z)$ and $h(z) = -\cos(2z)$, respectively. We then found the derivatives of the numerator and denominator using the product rule and chain rule of differentiation, respectively.

Q: What is the product rule of differentiation?

A: The product rule of differentiation states that if $f(z) = g(z)h(z)$, then $f'(z) = g'(z)h(z) + g(z)h'(z)$.

Q: How did you apply the product rule to the numerator of the function $G(z)$?

A: To apply the product rule to the numerator of the function $G(z)$, we identified the two functions as $g(z) = 5\sin(z)$ and $h(z) = \cos(z)$, respectively. We then found the derivatives of the two functions using the chain rule of differentiation.

Q: What is the chain rule of differentiation?

A: The chain rule of differentiation states that if $f(z) = g(h(z))$, then $f'(z) = g'(h(z))h'(z)$.

Q: How did you apply the chain rule to the denominator of the function $G(z)$?

A: To apply the chain rule to the denominator of the function $G(z)$, we identified the outer function as $g(z) = -\cos(2z)$ and the inner function as $h(z) = 2z$, respectively. We then found the derivatives of the outer and inner functions using the chain rule of differentiation.

In this article, we addressed some common questions and concerns related to the derivative of the function $G(z) = \frac{5}{\tan(z) - \cot(z)}$. We provided step-by-step explanations and examples to help clarify the concepts and procedures involved in finding the derivative of the function. We hope that this article has been helpful in providing a deeper understanding of the derivative of the function $G(z)$.