Calculus: Find The 51st Derivative At X=1

Welcome, math enthusiasts! Today, we're diving deep into the fascinating world of calculus to tackle a problem that might initially look intimidating: finding the 51st derivative of a given function and evaluating it at a specific point. Don't worry, we'll break it down step-by-step, making this complex task feel manageable and even enjoyable. Our primary goal is to determine the value of $\frac{d^{(51)} y}{d x^{(51)}}$ at $x=1$, where the function $y$ is defined by the differential equation $\frac{d y}{d x}=5 x^{43}+96 e^{x-1}$. This journey will not only test our understanding of differentiation rules but also highlight the power of recognizing patterns in derivatives.

Understanding the Foundation: Derivatives and Their Behavior

Before we embark on finding the 51st derivative, let's refresh our understanding of what derivatives represent and how they behave. The first derivative, $\frac{d y}{d x}$, tells us about the rate of change of a function. The second derivative, $\frac{d^2 y}{d x^2}$, describes the rate of change of the first derivative, essentially giving us information about the concavity of the original function. As we take successive derivatives, we are essentially peeling back layers of the function's behavior, revealing more intricate details about its shape and how it transforms.

In this problem, we are given $\frac{d y}{d x}=5 x^{43}+96 e^{x-1}$. This means our function $y$ is the antiderivative of this expression. However, the problem directly asks for the 51st derivative of $y$. This implies we need to differentiate the given expression for $\frac{d y}{d x}$ a total of 50 more times to reach the 51st derivative. Let's denote the given expression as $f(x) = 5 x^{43}+96 e^{x-1}$. We are looking for $\frac{d^{50}}{dx^{50}}[f(x)]$.

Deconstructing the Function: Polynomial and Exponential Parts

Our function, $f(x) = 5 x^{43}+96 e^{x-1}$, is composed of two distinct parts: a polynomial term, $5x^{43}$, and an exponential term, $96e^{x-1}$. To find the 51st derivative, we can analyze the derivatives of each part separately and then combine the results. This is a direct application of the linearity of differentiation, which states that the derivative of a sum is the sum of the derivatives, and the derivative of a constant times a function is the constant times the derivative of the function.

Let's first focus on the polynomial term: $5x^{43}$. Recall the power rule for differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$. Applying this repeatedly, we get:

• The 1st derivative of $5x^{43}$ is $5 imes 43 x^{42} = 215x^{42}$.
• The 2nd derivative is $215 imes 42 x^{41}$.
• ...and so on.

Notice a pattern here? With each differentiation, the power of $x$ decreases by 1, and a new coefficient is introduced. When we reach the 43rd derivative of $x^{43}$, the term will become a constant: $\frac{d^{43}}{dx^{43}}(x^{43}) = 43!$. For any derivative higher than the original power of $x$, the result will be zero. Since we are looking for the 50th derivative of $5x^{43}$ (to get to the 51st derivative of $y$), and the highest power of $x$ is 43, all derivatives of $5x^{43}$ from the 44th derivative onwards will be zero. Therefore, $\frac{d^{50}}{dx^{50}}(5x^{43}) = 0$.

Now, let's consider the exponential term: $96e^{x-1}$. The beauty of the exponential function $e^u$ is that its derivative is itself, $\frac{d}{du}(e^u) = e^u$. When we have a function of the form $e^{ax+b}$, its derivative is $a e^{ax+b}$. In our case, the exponent is $x-1$, so $a=1$ and $b=-1$. Thus, the derivative of $e^{x-1}$ is $1 imes e^{x-1} = e^{x-1}$.

This means that no matter how many times we differentiate $e^{x-1}$, the result will always be $e^{x-1}$. The same applies to $96e^{x-1}$; its first derivative is $96e^{x-1}$, its second derivative is $96e^{x-1}$, and its 50th derivative is also $96e^{x-1}$. So, $\frac{d^{50}}{dx^{50}}(96e^{x-1}) = 96e^{x-1}$.

Combining the Results: The 51st Derivative

Now, we can combine the derivatives of the two parts to find the 50th derivative of $f(x) = 5 x^{43}+96 e^{x-1}$.

$\frac{d^{50}}{dx^{50}}[f(x)] = \frac{d^{50}}{dx^{50}}(5x^{43}) + \frac{d^{50}}{dx^{50}}(96e^{x-1})$

Since we found that $\frac{d^{50}}{dx^{50}}(5x^{43}) = 0$ and $\frac{d^{50}}{dx^{50}}(96e^{x-1}) = 96e^{x-1}$, we have:

$\frac{d^{50}}{dx^{50}}[f(x)] = 0 + 96e^{x-1} = 96e^{x-1}$

This expression, $96e^{x-1}$, is the 50th derivative of the original function $\frac{d y}{d x}$. Therefore, it represents the 51st derivative of y with respect to x, $\frac{d^{(51)} y}{d x^{(51)}}$.

Evaluating at x=1: The Final Step

The problem asks us to find the value of this 51st derivative at $x=1$. So, we substitute $x=1$ into our result:

$\frac{d^{(51)} y}{d x^{(51)}} \bigg|_{x=1} = 96e^{1-1}$

Since $1-1 = 0$, and any non-zero number raised to the power of 0 is 1 (i.e., $e^0 = 1$), we get:

$96e^0 = 96 imes 1 = 96$

And there you have it! The value of the 51st derivative of $y$ at $x=1$ is 96.

A Deeper Look: The Power of Pattern Recognition in Calculus

This problem beautifully illustrates the importance of pattern recognition in calculus. When faced with higher-order derivatives, especially of polynomial and exponential functions, identifying recurring patterns can save a tremendous amount of computational effort. For polynomials, the derivatives eventually become zero. For the exponential function $e^{kx}$, the derivatives cycle through a constant multiple of the original function. Recognizing these behaviors allows us to quickly determine the result of differentiating many times without having to perform each individual step.

Consider the polynomial term $5x^{43}$. Its derivatives are:

• 1st: $5 imes 43 x^{42}$
• 2nd: $5 imes 43 imes 42 x^{41}$
• 3rd: $5 imes 43 imes 42 imes 41 x^{40}$
• ...and so on.

The general form of the $n$-th derivative of $cx^m$ (where $n eq m$) is c imes rac{m!}{(m-n)!} x^{m-n}. When $n > m$, the term $m-n$ becomes negative, and if we were to continue the pattern, we would eventually get a factor of $x$ raised to a negative power, but more importantly, the factorial term in the numerator rac{m!}{(m-n)!} becomes undefined in the standard sense, or if we think of it as $\frac{d^m}{dx^m}(x^m) = m!$, then $\frac{d^{m+1}}{dx^{m+1}}(x^m) = 0$. In our case, $m=43$, and we are looking for the 50th derivative. Since $50 > 43$, the derivative of the polynomial term is indeed zero.

For the exponential term $96e^{x-1}$, let $g(x) = 96e^{x-1}$.

• g'(x) = 96 imes rac{d}{dx}(e^{x-1}) = 96 imes e^{x-1} imes rac{d}{dx}(x-1) = 96e^{x-1} imes 1 = 96e^{x-1}.
• g''(x) = rac{d}{dx}(96e^{x-1}) = 96e^{x-1}.

This recursive property means that $\frac{d^n}{dx^n}(96e^{x-1}) = 96e^{x-1}$ for any positive integer $n$. This property is fundamental to understanding exponential functions and their role in differential equations.

The Significance of the Derivative at a Point

Evaluating the derivative at a specific point, like $x=1$ in this problem, provides a snapshot of the function's instantaneous rate of change at that exact location. For the 51st derivative, it tells us about the behavior of the function's