Answer :
Certainly! To find the derivative of the function [tex]\( y = x^{6x} \)[/tex] using logarithmic differentiation, we will follow a step-by-step approach.
### Step 1: Define the Function
We start with the function:
[tex]\[ y = x^{6x} \][/tex]
### Step 2: Take the Natural Logarithm on Both Sides
Applying the natural logarithm to both sides of the equation, we get:
[tex]\[ \ln(y) = \ln(x^{6x}) \][/tex]
### Step 3: Apply the Property of Logarithms
Using the logarithm property [tex]\(\ln(a^b) = b \ln(a)\)[/tex], the right-hand side becomes:
[tex]\[ \ln(y) = 6x \ln(x) \][/tex]
### Step 4: Differentiate Both Sides with Respect to [tex]\( x \)[/tex]
Use implicit differentiation to differentiate both sides with respect to [tex]\( x \)[/tex]. Remember the chain rule on the left side and product rule on the right side:
[tex]\[ \frac{d}{dx}[\ln(y)] = \frac{d}{dx}[6x \ln(x)] \][/tex]
### Step 5: Simplify the Derivative
The derivative of [tex]\( \ln(y) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( \frac{1}{y} y' \)[/tex]. On the right-hand side, we use the product rule for the expression [tex]\( 6x \ln(x) \)[/tex]:
[tex]\[ \frac{1}{y} y' = \frac{d}{dx}[6x \ln(x)] \][/tex]
Applying the derivative:
[tex]\[ \frac{1}{y} y' = 6 \frac{d}{dx}[x \ln(x)] \][/tex]
Using the product rule for [tex]\( x \ln(x) \)[/tex]:
[tex]\[ \frac{d}{dx}[x \ln(x)] = x \frac{d}{dx}[\ln(x)] + \ln(x) \frac{d}{dx}[x] \][/tex]
[tex]\[ \frac{d}{dx}[x \ln(x)] = x \cdot \frac{1}{x} + \ln(x) \cdot 1 \][/tex]
[tex]\[ \frac{d}{dx}[x \ln(x)] = 1 + \ln(x) \][/tex]
Therefore,
[tex]\[ \frac{1}{y} y' = 6 (1 + \ln(x)) \][/tex]
### Step 6: Solve for [tex]\( y' \)[/tex]
Now multiply both sides by [tex]\( y \)[/tex] to solve for [tex]\( y' \)[/tex]:
[tex]\[ y' = y \cdot 6 (1 + \ln(x)) \][/tex]
Substitute [tex]\( y = x^{6x} \)[/tex] back into the equation:
[tex]\[ y' = x^{6x} \cdot 6 (1 + \ln(x)) \][/tex]
### Final Answer
The derivative of the function [tex]\( y = x^{6x} \)[/tex] is:
[tex]\[ y' = 6 x^{6x} (1 + \ln(x)) \][/tex]
Hence,
[tex]\[ y'(x) = 6 x^{6x} (\ln(x) + 1) \][/tex]
### Step 1: Define the Function
We start with the function:
[tex]\[ y = x^{6x} \][/tex]
### Step 2: Take the Natural Logarithm on Both Sides
Applying the natural logarithm to both sides of the equation, we get:
[tex]\[ \ln(y) = \ln(x^{6x}) \][/tex]
### Step 3: Apply the Property of Logarithms
Using the logarithm property [tex]\(\ln(a^b) = b \ln(a)\)[/tex], the right-hand side becomes:
[tex]\[ \ln(y) = 6x \ln(x) \][/tex]
### Step 4: Differentiate Both Sides with Respect to [tex]\( x \)[/tex]
Use implicit differentiation to differentiate both sides with respect to [tex]\( x \)[/tex]. Remember the chain rule on the left side and product rule on the right side:
[tex]\[ \frac{d}{dx}[\ln(y)] = \frac{d}{dx}[6x \ln(x)] \][/tex]
### Step 5: Simplify the Derivative
The derivative of [tex]\( \ln(y) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( \frac{1}{y} y' \)[/tex]. On the right-hand side, we use the product rule for the expression [tex]\( 6x \ln(x) \)[/tex]:
[tex]\[ \frac{1}{y} y' = \frac{d}{dx}[6x \ln(x)] \][/tex]
Applying the derivative:
[tex]\[ \frac{1}{y} y' = 6 \frac{d}{dx}[x \ln(x)] \][/tex]
Using the product rule for [tex]\( x \ln(x) \)[/tex]:
[tex]\[ \frac{d}{dx}[x \ln(x)] = x \frac{d}{dx}[\ln(x)] + \ln(x) \frac{d}{dx}[x] \][/tex]
[tex]\[ \frac{d}{dx}[x \ln(x)] = x \cdot \frac{1}{x} + \ln(x) \cdot 1 \][/tex]
[tex]\[ \frac{d}{dx}[x \ln(x)] = 1 + \ln(x) \][/tex]
Therefore,
[tex]\[ \frac{1}{y} y' = 6 (1 + \ln(x)) \][/tex]
### Step 6: Solve for [tex]\( y' \)[/tex]
Now multiply both sides by [tex]\( y \)[/tex] to solve for [tex]\( y' \)[/tex]:
[tex]\[ y' = y \cdot 6 (1 + \ln(x)) \][/tex]
Substitute [tex]\( y = x^{6x} \)[/tex] back into the equation:
[tex]\[ y' = x^{6x} \cdot 6 (1 + \ln(x)) \][/tex]
### Final Answer
The derivative of the function [tex]\( y = x^{6x} \)[/tex] is:
[tex]\[ y' = 6 x^{6x} (1 + \ln(x)) \][/tex]
Hence,
[tex]\[ y'(x) = 6 x^{6x} (\ln(x) + 1) \][/tex]
