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Write derivative formulas for the functions. $$ f(x)=\frac{12.6\left(4.8^{x}\right)}{x^{2}} $$

Short Answer

Expert verified
The derivative is \( \frac{12.6 \cdot 4.8^x \cdot (x^2 \ln(4.8) - 2x)}{x^4} \).

Step by step solution

01

Identify Derivative Needed

This function is a quotient of a constant times an exponential function and a power of x. This means we'll use the quotient rule for derivatives: if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of \( \frac{u}{v} \) is \( \frac{u'v - uv'}{v^2} \). For this problem, \( u(x) = 12.6 \times 4.8^x \) and \( v(x) = x^2 \).
02

Derivative of the Numerator

Find the derivative of the numerator function \( u(x) = 12.6 \times 4.8^x \). The derivative of an exponential function \( a^x \) is \( a^x \ln(a) \). Thus, \( u'(x) = 12.6 \times 4.8^x \ln(4.8) \).
03

Derivative of the Denominator

Now, find the derivative of the denominator function \( v(x) = x^2 \). The derivative of \( x^n \) is \( nx^{n-1} \), so \( v'(x) = 2x \).
04

Apply the Quotient Rule

Apply the quotient rule: \( \frac{u'v - uv'}{v^2} \). Substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into the formula. \[ (12.6 \times 4.8^x \ln(4.8)) \times (x^2) - (12.6 \times 4.8^x) \times (2x) \over (x^2)^2 \]
05

Simplify the Expression

Simplify the expression obtained in Step 4:\[ \frac{12.6 \times 4.8^x(x^2 \ln(4.8) - 2x)}{x^4} \] This is the derivative \( f'(x) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
Understanding the quotient rule is essential when dealing with a function that is a fraction, like our given function. The quotient rule helps us find the derivative of a quotient of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), and these are arranged in a ratio \( \frac{u(x)}{v(x)} \), the derivative of this ratio is given by the quotient rule:\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \]This formula might seem complex, but it's just a process to find out how fast one function is changing relative to another. Here's a simple breakdown:
  • \( u' \) is the derivative of the numerator.
  • \( v \) is the original denominator function.
  • \( u \) is the original numerator function.
  • \( v' \) is the derivative of the denominator.
Calculate each component, follow the formula, and you'll have the derivative of the entire fraction.
Exponential Functions
Exponential functions are functions where the variable appears in the exponent. The most common form is \( a^x \), where \( a \) is a positive constant. In our function, \( u(x) = 12.6 \times 4.8^x \), the exponential part is \( 4.8^x \).Exponential functions grow or decay rapidly, depending on the value of \( a \):
  • If \( a > 1 \), the function grows as \( x \) increases.
  • If \( 0 < a < 1 \), the function decays as \( x \) increases.
One of the crucial properties of exponential functions is their derivative. The derivative of \( a^x \) is \( a^x \ln(a) \). For our function, this means the derivative of \( 4.8^x \) is \( 4.8^x \ln(4.8) \). So when differentiating \( u(x) \), you multiply this result by the constant outside the exponential, giving us \( 12.6 \times 4.8^x \ln(4.8) \).
Derivative Formulas
In calculus, derivative formulas are fundamental tools for finding how functions change. Derivatives give us the rate of change or the slope of the function at any point. Here are some basic derivatives to remember:
  • The derivative of \( c \cdot f(x) \), where \( c \) is a constant, is \( c \cdot f'(x) \).
  • The derivative of \( x^n \) is \( nx^{n-1} \), called the power rule.
  • The derivative of the exponential function \( a^x \) is \( a^x \ln(a) \).
  • When functions are combined like in sums or differences, derivatives can be added or subtracted as \( (f+g)' = f' + g' \).
Using these formulas, we can tackle complex functions by breaking them down into components and applying appropriate rules. Our task can be split into these elements for effective differentiation by applying the quotient rule and exponential/ power rule as needed.
Power Rule
The power rule is one of the simplest yet most powerful tools in calculus for finding the derivative of a function. Whenever you have a function of the form \( x^n \), the derivative is \( nx^{n-1} \). This rule assumes that \( n \) can be any real number.Let's look at how this rule is applied: In our function, the denominator \( v(x) = x^2 \) is an example where the power rule is useful. According to the power rule:If \( v(x) = x^2 \), then \( v'(x) = 2x \).The power rule makes finding derivatives straightforward, and when used in combination with other rules, it allows us to handle a variety of complex functions. This rule applies not just for whole numbers but also for fractions, negatives, and decimals as exponents. Always reduce the power by one and bring the original power down as a coefficient.

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