Derivatives: Composite Function & Quotient Rule

Hey guys! Today, we're diving into the exciting world of calculus, tackling two common derivative problems. We'll explore how to find the derivative of a composite function and how to handle the derivative of a quotient. Let's break it down step by step, making sure everyone understands the process.

Derivative of a Composite Function: $cot^{15}(x)$

Okay, so our first task is to find the derivative of $\operatorname{ctg}^{15} x$. This looks a bit intimidating, but don't worry, we'll use the chain rule to simplify it. The chain rule is your best friend when dealing with composite functions – functions within functions. In this case, we have the cotangent function raised to the power of 15.

To apply the chain rule, we need to identify the outer and inner functions. Here, the outer function is $u^{15}$, where $u = \operatorname{ctg} x$ is the inner function. The chain rule states that if we have a composite function $y = f(g(x))$, then its derivative is $y' = f'(g(x)) \\cdot g'(x)$.

Let's find the derivative of the outer function, $u^{15}$, with respect to $u$. Using the power rule, which states that the derivative of $x^n$ is $nx^{n-1}$, we get:

$\frac{d}{du}(u^{15}) = 15u^{14}$

Now, we need to find the derivative of the inner function, $u = \operatorname{ctg} x$, with respect to $x$. Recall that the derivative of $\operatorname{ctg} x$ is $-\csc^2 x$. So, we have:

$\frac{d}{dx}(\operatorname{ctg} x) = -\csc^2 x$

Now, we apply the chain rule by multiplying the derivative of the outer function (evaluated at the inner function) by the derivative of the inner function:

$\frac{d}{dx}(\operatorname{ctg}^{15} x) = 15(\operatorname{ctg} x)^{14} \\cdot (-\csc^2 x) = -15 \operatorname{ctg}^{14} x \csc^2 x$

So, the derivative of $\operatorname{ctg}^{15} x$ is $-15 \operatorname{ctg}^{14} x \csc^2 x$. Remember, the key to the chain rule is to break down the composite function into its individual parts, find their derivatives, and then multiply them together. This might seem complex at first, but with practice, it becomes second nature. Just keep breaking down those functions and applying the rule step by step.

Derivative of a Quotient: $\frac{x^2-5x+6}{x+1}$

Next up, we need to find the derivative of the quotient $\frac{x^2-5x+6}{x+1}$. For this, we'll use the quotient rule. The quotient rule is used when you have a function that is the ratio of two other functions. The rule states that if we have a function $y = \frac{u(x)}{v(x)}$, then its derivative is $y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

In our case, $u(x) = x^2 - 5x + 6$ and $v(x) = x + 1$. First, let's find the derivatives of $u(x)$ and $v(x)$.

The derivative of $u(x) = x^2 - 5x + 6$ is found using the power rule and the sum/difference rule. The power rule states that the derivative of $x^n$ is $nx^{n-1}$, and the sum/difference rule states that the derivative of a sum or difference of terms is the sum or difference of their derivatives. So, we have:

$u'(x) = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5$

The derivative of $v(x) = x + 1$ is straightforward:

$v'(x) = \frac{d}{dx}(x + 1) = 1$

Now, we apply the quotient rule:

$\frac{d}{dx} \left( \frac{x^2-5x+6}{x+1} \right) = \frac{(2x - 5)(x + 1) - (x^2 - 5x + 6)(1)}{(x + 1)^2}$

Next, we need to simplify the expression. First, expand the terms in the numerator:

$(2x - 5)(x + 1) = 2x^2 + 2x - 5x - 5 = 2x^2 - 3x - 5$

Now, subtract $(x^2 - 5x + 6)$ from the expanded term:

$2x^2 - 3x - 5 - (x^2 - 5x + 6) = 2x^2 - 3x - 5 - x^2 + 5x - 6 = x^2 + 2x - 11$

So, the derivative simplifies to:

$\frac{d}{dx} \left( \frac{x^2-5x+6}{x+1} \right) = \frac{x^2 + 2x - 11}{(x + 1)^2}$

Therefore, the derivative of $\frac{x^2-5x+6}{x+1}$ is $\frac{x^2 + 2x - 11}{(x + 1)^2}$. Remember, guys, the quotient rule might look complicated, but it’s just about carefully applying the formula and then simplifying the result. Practice makes perfect, so keep at it!

Key Takeaways

Let's recap what we've learned today:

• Chain Rule: Used for composite functions. Break down the function into inner and outer parts, find their derivatives, and multiply them together.
• Quotient Rule: Used for functions that are ratios. Apply the formula carefully and simplify the result.

Both rules are fundamental in calculus, and mastering them will greatly improve your ability to solve derivative problems. Keep practicing, and you'll become a pro in no time!

Understanding derivatives is crucial in calculus, especially when dealing with complex functions. The chain rule and quotient rule are two essential tools in your mathematical toolkit. By mastering these rules, you can tackle a wide range of problems involving composite functions and quotients. Remember, the key is to break down complex problems into smaller, manageable steps, and to practice consistently. This approach not only helps in understanding the concepts but also builds confidence in solving more challenging problems.