Sure, let's find the derivative [tex]\(\frac{dy}{dx}\)[/tex] for the equation [tex]\(y = x^y\)[/tex] using implicit differentiation step-by-step.
### Step 1: Rewrite the Equation
Let's start by taking the natural logarithm of both sides of the equation to make differentiation easier:
[tex]\[ \ln(y) = \ln(x^y) \][/tex]
### Step 2: Apply Logarithm Properties
We can use the property of logarithms that states [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[ \ln(y) = y \ln(x) \][/tex]
### Step 3: Differentiate Both Sides with Respect to [tex]\(x\)[/tex]
We will use implicit differentiation here. Differentiate both sides with respect to [tex]\(x\)[/tex]:
#### Left Side:
[tex]\[ \frac{d}{dx} [\ln(y)] = \frac{1}{y} \frac{dy}{dx} \][/tex]
#### Right Side:
Use the product rule for differentiation:
[tex]\[ \frac{d}{dx} [y \ln(x)] = \ln(x) \frac{dy}{dx} + y \frac{d}{dx} [\ln(x)] \][/tex]
Since [tex]\(\frac{d}{dx} [\ln(x)] = \frac{1}{x}\)[/tex], the right side becomes:
[tex]\[ \ln(x) \frac{dy}{dx} + y \frac{1}{x} \][/tex]
### Step 4: Set Equations Equal
Combine the differentiated left and right sides:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \ln(x) \frac{dy}{dx} + \frac{y}{x} \][/tex]
### Step 5: Solve for [tex]\(\frac{dy}{dx}\)[/tex]
Isolate [tex]\(\frac{dy}{dx}\)[/tex] on one side of the equation:
[tex]\[ \frac{1}{y} \frac{dy}{dx} - \ln(x) \frac{dy}{dx} = \frac{y}{x} \][/tex]
Factor out [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \left( \frac{1}{y} - \ln(x) \right) \frac{dy}{dx} = \frac{y}{x} \][/tex]
Thus,
[tex]\[ \frac{dy}{dx} = \frac{\frac{y}{x}}{\frac{1}{y} - \ln(x)} \][/tex]
### Step 6: Simplify
Simplifying the equation gives us:
[tex]\[ \frac{dy}{dx} = \frac{y}{x \left( \frac{1}{y} - \ln(x) \right)} \][/tex]
Combining the terms in the denominator:
[tex]\[ \frac{dy}{dx} = \frac{y^2}{x (1 - y \ln(x))} \][/tex]
So, the derivative [tex]\(\frac{dy}{dx}\)[/tex] for the equation [tex]\( y = x^y \)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{y^2}{x (1 - y \ln(x))} \][/tex]
