1. If X = E^y, Show That Dy/dx = 1/x. 2. Find Dy/dx If X = E^t, Y = E^(e^t). 3. Find Dy/dx If Y = X^5 + 5. 4. If X^(2/3) + Y^(2/3) = (x+y)^(2/3), Find Dy/dx.

by ADMIN 158 views

Calculus, a cornerstone of mathematical analysis, equips us with the tools to understand rates of change and accumulation. Among the fundamental concepts in calculus, derivatives play a pivotal role in describing the instantaneous rate of change of a function. This article delves into various derivative problems, focusing on exponential and logarithmic functions, and provides step-by-step solutions and explanations. Understanding derivatives is crucial for various applications in science, engineering, economics, and computer science.

1. Differentiating Logarithmic and Exponential Functions

In this section, we tackle the problem: if x = ey**, show that dy/dx = log x / (1 + log x)². This problem elegantly combines exponential and logarithmic functions, showcasing the inverse relationship between them. To solve this, we must first express y in terms of x, then differentiate. Let's break down the process:

Step 1: Express y in terms of x

Given x = ey**, we need to isolate y. This is achieved by taking the natural logarithm (ln) of both sides. The natural logarithm is the inverse function of the exponential function with base e. Thus:

ln(x) = ln(e*y)

Using the property of logarithms that ln(a*b) = b ln(a), we get:

ln(x) = y ln(e)

Since ln(e) = 1, we simplify to:

y = ln(x)

This crucial step rewrites the equation in a form that allows us to directly differentiate y with respect to x.

Step 2: Differentiate y with respect to x

Now that we have y = ln(x), we can find dy/dx. However, the expression we need to arrive at involves log x (base 10 logarithm), not ln x (natural logarithm). We need to either remember the derivative of log base a of x, or we need to convert the natural logarithm to a base-10 logarithm, using the change of base formula:

ln(x) = log(x) / log(e)

Substituting this back into our expression for y:

y = log(x) / log(e)

Now, we differentiate both sides with respect to x:

dy/dx = d/dx [log(x) / log(e)]

Since 1/log(e) is a constant, we can take it out of the derivative:

dy/dx = (1 / log(e)) * d/dx [log(x)]

The derivative of log(x) (base 10) is 1 / (x ln(10)), so:

dy/dx = (1 / log(e)) * (1 / (x ln(10)))

To simplify, we can use the change of base formula in reverse. Specifically, 1/log(e) = ln(10), so:

dy/dx = ln(10) / (x ln(10))

Now we encounter a roadblock. It seems that something has gone wrong as we are not arriving at the expression provided, which involves (1 + log x) in the denominator. Let's revert to differentiating y = ln(x) directly and see if we can manipulate the result. We know that:

dy/dx = d/dx [ln(x)] = 1/x

Now, we need to algebraically manipulate this expression to match the target expression: log x / (1 + log x)². This requires a clever substitution. Remembering x = e*y, we know that ln(x) = y. However, this does not seem to directly lead us to the desired expression. Perhaps the problem statement has a mistake. Let's try an alternative approach.

Instead of substituting early, let us try implicit differentiation on the original equation x = e^y. Differentiating both sides with respect to x, we get:

1 = e^y * dy/dx

Solving for dy/dx:

dy/dx = 1 / e^y

Since x = e^y, we have:

dy/dx = 1/x

Now, we know that y = ln(x). Let's see if we can rewrite 1/x in terms of log(x). We know that x = 10^log(x). So, 1/x = 1/10^log(x). This doesn't seem to directly lead to the provided solution either.

Conclusion: There appears to be an error in the problem statement. The derivative of y = ln(x) is 1/x, which can be written as 1/ey using the given substitution. The expression log x / (1 + log x)² does not seem to be the correct derivative based on the given information.

2. Finding Derivatives of Composite Exponential Functions

Next, consider the problem: Find dy/dx if x = et**, y = eet***.** This involves finding the derivative of a composite function, which requires the chain rule. The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. It states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). Let's apply this to our problem.

Step 1: Identify the functions and their derivatives

We have two equations: x = et and y = eet***. We need to find dy/dx. Since both x and y are expressed in terms of t, we can use the chain rule in the form:

dy/dx = (dy/dt) / (dx/dt)

First, let's find dx/dt:

dx/dt = d/dt (et) = et*

Next, let's find dy/dt. Here, we have a composite function. Let u = et. Then y = eu. So:

dy/dt = dy/du * du/dt

dy/du = d/du (eu) = eu*

du/dt = d/dt (et) = et*

Therefore:

dy/dt = eu** * et = eet* * et

Step 2: Apply the chain rule

Now we have dx/dt = et** and dy/dt = eet * et. Applying the chain rule:

dy/dx = (dy/dt) / (dx/dt) = (eet * et) / (et)*

Simplifying, we get:

dy/dx = eet*

This is the derivative of y with respect to x. The use of the chain rule is crucial in solving this type of problem, allowing us to break down a complex derivative into simpler components.

3. Basic Power Rule Differentiation

Consider the problem: Find dy/dx if y = x⁵ + 5. This is a straightforward application of the power rule and the constant rule of differentiation. The power rule states that if y = xn, then dy/dx = n xn-1**. The constant rule states that the derivative of a constant is zero.

Step 1: Apply the power rule and constant rule

We have y = x⁵ + 5. Differentiating with respect to x:

dy/dx = d/dx (x⁵ + 5)

Using the sum rule of differentiation, which states that d/dx (u + v) = d/dx (u) + d/dx (v):

dy/dx = d/dx (x⁵) + d/dx (5)

Applying the power rule to the first term:

d/dx (x⁵) = 5 x⁴

Applying the constant rule to the second term:

d/dx (5) = 0

Step 2: Combine the results

Therefore:

dy/dx = 5x⁴ + 0 = 5x⁴**

This simple example demonstrates the application of basic differentiation rules. Understanding these rules is fundamental for tackling more complex derivative problems.

4. Implicit Differentiation and Homogeneous Functions

Lastly, let's tackle the problem: If x⅔ + y⅔ = (x + y)⅔, then show that dy/dx is something specific (The target expression was not provided in the original question). This problem involves implicit differentiation and the concept of homogeneous functions. Implicit differentiation is used when y cannot be easily expressed as an explicit function of x. We differentiate both sides of the equation with respect to x, treating y as a function of x and using the chain rule when necessary.

Step 1: Implicit Differentiation

Given the equation x⅔ + y⅔ = (x + y)⅔, we differentiate both sides with respect to x:

d/dx (x⅔ + y⅔) = d/dx ((x + y)⅔)

Using the sum rule:

d/dx (x⅔) + d/dx (y⅔) = d/dx ((x + y)⅔)

Applying the power rule and chain rule:

(⅔) x(-⅓) + (⅔) y(-⅓) * (dy/dx) = (⅔) (x + y)(-⅓) * (1 + dy/dx)

Step 2: Simplify and solve for dy/dx

Now, we need to simplify the equation and solve for dy/dx:

(⅔) x(-⅓) + (⅔) y(-⅓) * (dy/dx) = (⅔) (x + y)(-⅓) + (⅔) (x + y)(-⅓) * (dy/dx)

Divide both sides by ⅔:

x(-⅓) + y(-⅓) * (dy/dx) = (x + y)(-⅓) + (x + y)(-⅓) * (dy/dx)

Rearrange terms to isolate dy/dx:

y(-⅓) * (dy/dx) - (x + y)(-⅓) * (dy/dx) = (x + y)(-⅓) - x(-⅓)

Factor out dy/dx:

(dy/dx) [y(-⅓) - (x + y)(-⅓)] = (x + y)(-⅓) - x(-⅓)

Solve for dy/dx:

dy/dx = [(x + y)(-⅓) - x(-⅓)] / [y(-⅓) - (x + y)(-⅓)]

This is the expression for dy/dx. We could further manipulate this algebraically, but without the target expression, it's difficult to determine the simplest form.

Note: If we had to simplify this to dy/dx = -√(y/x), we would multiply numerator and denominator by (x+y)(1/3)x(1/3)y^(1/3) to clear negative exponents, and then try to factor perfect squares. The original problem statement likely omitted the target expression for simplification, which is a common practice in calculus exercises.

Conclusion

This article explored various derivative problems, encompassing logarithmic, exponential, and algebraic functions. We utilized fundamental rules of differentiation, such as the power rule, constant rule, chain rule, and implicit differentiation. While one of the initial problems appears to have a discrepancy, the overall exploration provides a solid foundation for understanding and applying derivatives in calculus. Derivatives are a powerful tool for analyzing rates of change and are essential for numerous applications across diverse fields. The ability to differentiate various types of functions is a core skill for anyone working in mathematics, science, or engineering.