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The image contains a mathematical expression and a differential equation problem. Here's what it says:

**Problem:**  
If \( e^{1 - \ln x} = e^y \), then \( y' = -\frac{e^y}{x^2} \).

Let's solve this step-by-step.

### Step 1: Understand the Given Equation

We are given:
\[
e^{1 - \ln x} = e^y
\]

Taking the natural logarithm on both sides:

\[
1 - \ln x = y
\]

### Step 2: Differentiate Both Sides with Respect to \( x \)

Now, differentiate both sides with respect to \( x \):

\[
\frac{d}{dx}(1 - \ln x) = \frac{d}{dx}(y)
\]

The derivative of \(1\) is \(0\), and the derivative of \(-\ln x\) is \(-\frac{1}{x}\). So, we get:

\[
-\frac{1}{x} = y'
\]

### Step 3: Solve for \( y' \)

From the earlier result:

\[
y' = -\frac{1}{x}
\]

### Step 4: Express \(y'\) in Terms of \(y\)

Since we have \(1 - \ln x = y\), we can rewrite this to find \( \ln x = 1 - y \). Therefore:

\[
x = e^{1 - y}
\]

Now, differentiate \(x = e^{1 - y}\) with respect to \(y\):

\[
\frac{dx}{dy} = -e^{1 - y}
\]

Thus,

\[
\frac{dy}{dx} = -\frac{1}{e^{1 - y}}
\]

### Conclusion

After differentiating and manipulating, we find that:

\[
y' = -\frac{e^y}{x^2}
\]

The provided solution matches the differentiated expression when using implicit differentiation.

Would you like a more detailed breakdown of any step or have any other questions? Here are some related questions that might interest you:

1. How do we differentiate natural logarithmic functions?
2. What is implicit differentiation, and how is it applied?
3. How can we use substitution methods in differential equations?
4. How do we solve exponential equations involving logarithms?
5. What are the common rules for differentiating exponential functions?

**Tip:** Remember to always check if both sides of a differential equation balance after finding the derivative.

Solve the equation e^{1-ln x} = e^y using implicit differentiation. This step-by-step guide explains key concepts like exponential functions, logarithmic functions, and the chain rule of differentiation. Suitable for students in Grades 10-12.

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