Question

Given that $q(x)=\frac{5x^8+6x^5+5x^4+3x^2+20x+100}{10x^{10}+8x^9+6x^5+6x^2+4x+2},$ find $q^{\prime }(0)$ without computing $q^{\prime }(x).$ (Hint: Evaluate $f(0),f^{\prime }(0),g(0)$ and $g^{\prime }(0)$ where $f$ is the numerator of $q$ and $g$ is the denominator of $q.)$

Short answer

Using the Quotient Rule and evaluating the given functions at x = 0, we found that the value of q'(0) is -90.

Step-by-step solution

Evaluate $f(x)$ at $0$

We are given $f(x)=5x^8+6x^5+5x^4+3x^2+20x+100$. To find f(0), let x = 0: $f(0)=5(0{)}^8+6(0{)}^5+5(0{)}^4+3(0{)}^2+20(0)+100=100$

Evaluate $g(x)$ at $0$

We have $g(x)=10x^{10}+8x^9+6x^5+6x^2+4x+2$. To find g(0), let x = 0: $g(0)=10(0{)}^{10}+8(0{)}^9+6(0{)}^5+6(0{)}^2+4(0)+2=2$

Compute the derivative of $f(x)$ and evaluate it at $0$

To find $f^{\prime }(0)$, first compute the derivative of $f(x)$: $f^{\prime }(x)=40x^7+30x^4+20x^3+6x+20$ Now, evaluate $f^{\prime }(0)$: $f^{\prime }(0)=40(0{)}^7+30(0{)}^4+20(0{)}^3+6(0)+20=20$

Compute the derivative of $g(x)$ and evaluate it at $0$

To find $g^{\prime }(0)$, first compute the derivative of $g(x)$: $g^{\prime }(x)=100x^9+72x^8+30x^4+12x+4$ Now, evaluate $g^{\prime }(0)$: $g^{\prime }(0)=100(0{)}^9+72(0{)}^8+30(0{)}^4+12(0)+4=4$

Apply the Quotient Rule and plug in all values

Now, we can apply the Quotient Rule for finding $q^{\prime }(0)$: $q^{\prime }(0)=\frac{f^{\prime }(0)g(0)-f(0)g^{\prime }(0)}{g(0{)}^2}=\frac{20\cdot 2-100\cdot 4}{2^2}=\frac{40-400}{4}=\frac{-360}{4}=-90$ Therefore, the value of $q^{\prime }(0)$ is $-90$.

Key concepts

Quotient Rule

The Quotient Rule is a formula for taking the derivative of a quotient of two functions. It is useful when the function you're trying to differentiate, let's say $q(x)$, is a fraction where both the numerator, $f(x)$, and the denominator, $g(x)$, are also functions of $x$. The rule states:
$$q^{\prime }(x)=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g(x{)}^2}$$
This means that the derivative of the fraction is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all over the denominator squared. Remember that the process of differentiation tells us about the rate at which a function is changing at any given point. When applying this formula, always consider simplifying the function if possible before using the Quotient Rule to make the calculation easier.

Polynomial Functions

Polynomial functions are algebraic expressions that consist of terms in the form of $a_n{x}^n$, where $a_n$ is a coefficient, $x$ is the variable, and $n$ is a non-negative integer called the degree of the term. The highest degree of any term in the polynomial determines the degree of the polynomial function itself.
In our exercise, $f(x)$ and $g(x)$ are both polynomial functions. Evaluating these at $x=0$ simplifies the process because any term with an $x$ will become zero, leaving only the constant term. This is why $f(0)=100$ and $g(0)=2$ in our problem – all the terms with $x$ drop out. Polynomial functions are foundational in calculus, as they provide straightforward examples for the rules of differentiation and integration.

Function Derivatives

Function derivatives represent the crux of calculus. They provide instantaneous rates of change and slopes of tangent lines to functions at any given point. When we talk about the derivative of a function, we're looking for a new function, typically denoted as $f^{\prime }(x)$, which gives us this rate of change. For polynomial functions, like $f(x)$ and $g(x)$ in our exercise, deriving is mostly a matter of applying the power rule, which is to multiply the coefficient by the exponent and subtract one from the exponent.
For example, the derivative of $5x^8$ is $40x^7$. We use these derivatives in the Quotient Rule to find the derivative of a ratio of two functions, which was precisely our task in finding $q^{\prime }(0)$ for the given function.

Evaluating Functions at a Point

Evaluating functions at a specific point involves substituting the value of that point into the function. For polynomials, as mentioned, this often simplifies greatly when the point is zero, as all terms with $x$ subtract out. This is an essential skill in calculus, as it allows us to find the exact value of a function or its derivatives at particular points. These values can tell us many things about a function's behavior - such as whether it's increasing or decreasing, what the slope is at that point, and where it intercepts the y-axis.
In our practice problem, evaluating at zero, $x=0$, was crucial in simplifying the steps to find $q^{\prime }(0)$. Understanding how to do this effectively can make dealing with complex derivatives much more manageable.