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Last updated on September 1, 2025

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Derivative of e^y

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We use the derivative of e^y, which is e^y(dy/dx) when y is a function of x, as a tool to understand how the function changes with respect to a variable. Derivatives are crucial in various real-life applications including scientific modeling and economics. We will now discuss the derivative of e^y in detail.

Derivative of e^y for US Students
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What is the Derivative of e^y?

The derivative of ey is an important concept in calculus.

It is generally represented as d/dx(ey) or (ey)'.

Since e^y is an exponential function, its derivative is itself multiplied by the derivative of its exponent. The key concepts involved are:

Exponential Function: ey where e is a constant approximately equal to 2.71828.

Chain Rule: Used for differentiating composite functions like ey when y is a function of x. Implicit

Differentiation: Necessary when y is implicitly defined in terms of x.

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Derivative of e^y Formula

The derivative of ey, when y is a function of x, can be denoted as d/dx(ey) or (ey)'.

The formula used to differentiate ey is: d/dx(ey) = ey(dy/dx)

This formula is applicable whenever y is a differentiable function of x.

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Proofs of the Derivative of e^y

We can derive the derivative of e^y using different methods.

To show this, we will apply the chain rule and the concept of implicit differentiation.

Here are the methods used to prove this:

Using Chain Rule To prove the differentiation of ey using the chain rule, consider y as a function of x:

Let u = y, where y is a function of x.

Then, ey = eu.

Applying the chain rule, we get: d/dx(ey) = d/dx(eu) = eu * du/dx Since u = y, du/dx = dy/dx.

Therefore, d/dx(ey) = ey * dy/dx.

Using Implicit Differentiation Assume y is implicitly defined in terms of x.

Differentiate both sides of the equation ey = z with respect to x: d/dx(ey) = d/dx(z)

Using the chain rule, we get: ey * dy/dx = dz/dx

If z = e^y, then dz/dx = ey * dy/dx.

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Higher-Order Derivatives of e^y

When a function is differentiated multiple times, the resulting derivatives are known as higher-order derivatives.

Higher-order derivatives can provide more insight into the behavior of the function.

Consider the case of ey:

First Derivative: d/dx(ey) = ey(dy/dx)

Second Derivative: Differentiate the first derivative again, using the product rule: d²/dx²(ey) = d/dx(ey(dy/dx)) = (ey(dy/dx))(dy/dx) + ey(d²y/dx²)

This process can be continued for higher-order derivatives.

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Special Cases:

When y is constant, the derivative ey is simply ey multiplied by 0, which is 0. When y is a linear function of x, say y = ax, the derivative simplifies to ey * a.

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Common Mistakes and How to Avoid Them in Derivatives of e^y

Students often make errors when differentiating ey, especially when y is a function of x. These mistakes can be corrected by understanding the correct process. Here are some common mistakes and how to solve them:

Mistake 1

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Neglecting the Chain Rule

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Students may forget to apply the chain rule, leading to incorrect results. Always remember to multiply by dy/dx when differentiating ey where y is a function of x.

Mistake 2

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Ignoring Implicit Differentiation

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Students sometimes overlook the need for implicit differentiation, especially when y is not explicitly given as a function of x. Make sure to account for dy/dx.

Mistake 3

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Misapplying Product Rule

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In cases where ey is part of a product, students might incorrectly apply the product rule. Ensure proper use of the product rule, differentiating each component correctly.

Mistake 4

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Forgetting Constants in Exponential Functions

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When dealing with expressions like e(y+z), students often forget to differentiate all variables involved. Remember to apply the chain rule to each term.

Mistake 5

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Incorrectly Handling Higher-Order Derivatives

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Mistakes occur when calculating higher-order derivatives. Ensure each differentiation step is performed accurately using applicable rules.

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Examples Using the Derivative of e^y

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Problem 1

Calculate the derivative of e^(2y) with respect to x, given y = x² + 1.

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Here, we have f(y) = e(2y) and y = x² + 1.

Using the chain rule: d/dx(e(2y)) = e(2y) * d/dx(2y)

Since y = x² + 1, we have: d/dx(2y) = 2 * d/dx(y) = 2 * 2x = 4x

So, the derivative is: f'(x) = e(2y) * 4x = 4x * e(2(x²+1))

Thus, the derivative of e(2y) with respect to x is 4x * e(2(x²+1)).

Explanation

We differentiate e(2y) using the chain rule. By substituting y = x² + 1 and differentiating, we obtain the derivative.

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Problem 2

A company observes growth modeled by e^y, where y is the profit function y = 3x + 2. Find the rate of change of growth at x = 1.

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We have ey as the growth model and y = 3x + 2.

First, differentiate y: dy/dx = 3

Now, differentiate e^y: d/dx(ey) = ey * dy/dx = ey * 3

At x = 1, y = 3(1) + 2 = 5

Therefore, the rate of change of growth is: 3 * e5

Explanation

The company's growth is modeled by ey, where y is profit. The rate of change is found by differentiating and substituting x = 1 into the equation.

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Problem 3

Find the second derivative of e^y where y = ln(x).

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First, find the first derivative: d/dx(ey) = ey * dy/dx

Since y = ln(x), dy/dx = 1/x.

Therefore, d/dx(e^y) = ey * 1/x

Now, to find the second derivative: d²/dx²(ey) = d/dx(ey * 1/x)

Using the product rule: = (1/x) * d/dx(ey) + ey * d/dx(1/x) = (1/x) * (ey * 1/x) + ey * (-1/x²) = ey/x² - ey/x² = 0

So, the second derivative is 0.

Explanation

We start with the first derivative using implicit differentiation and apply the product rule for the second derivative. The terms simplify to zero.

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Problem 4

Show that d/dx(e^(3y)) = 3e^(3y) * dy/dx.

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Using the chain rule:

Consider f(x) = e(3y) d/dx(e(3y)) = e(3y) * d/dx(3y)

Since d/dx(3y) = 3 * dy/dx,

We have: d/dx(e(3y)) = 3e(3y) * dy/dx

Hence proved.

Explanation

We use the chain rule to differentiate e(3y). The derivative of the exponent is multiplied by the derivative of y, confirming the result.

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Problem 5

Differentiate e^(y/x) with respect to x.

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Use the chain rule and quotient rule:

d/dx(e(y/x)) = e(y/x) * d/dx(y/x)

Apply the quotient rule: d/dx(y/x) = (x * dy/dx - y * 1)/x² = (x * dy/dx - y)/x²

Therefore, the derivative is: e(y/x) * ((x * dy/dx - y)/x²)

Explanation

We apply the chain rule and quotient rule to find the derivative of e(y/x) by differentiating the exponential function and its exponent.

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FAQs on the Derivative of e^y

1.What is the derivative of e^y?

The derivative of ey with respect to x is ey * dy/dx, assuming y is a function of x.

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2.How is the derivative of e^y used in real life?

The derivative of ey is used in various fields such as physics for modeling exponential growth or decay, in economics for growth rates, and in engineering for signal processing.

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3.Can we differentiate e^y when y is a constant?

Yes, if y is a constant, d/dx(ey) = 0 because the derivative of a constant is zero.

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4.What rule is used to differentiate e^(y/x)?

We use both the chain rule and the quotient rule to differentiate e(y/x).

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5.Are the derivatives of e^y and e^x the same?

No, they are different. The derivative of ey is ey * dy/dx when y is a function of x, while the derivative of ex is simply ex.

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Important Glossaries for the Derivative of e^y

  • Derivative: A measure of how a function changes as its input changes, often represented as the slope of the tangent line to the curve of the function.

 

  • Exponential Function: A function of the form ey where e is the base of the natural logarithm, approximately equal to 2.71828.

 

  • Chain Rule: A fundamental rule in calculus used to differentiate compositions of functions.

 

  • Implicit Differentiation: A method to find derivatives of functions that are not explicitly solved for one variable in terms of another.

 

  • Quotient Rule: A technique for differentiating functions that are the ratio of two differentiable functions. 
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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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