Answered

Use logarithmic differentiation to find:

[tex]\[ \frac{dy}{dx} \quad \text{if} \quad y = e^{x^2} \tan^3(x) \][/tex]



Answer :

To find the derivative of [tex]\( y = e^{x^2} \tan^3(x) \)[/tex] using logarithmic differentiation, follow these steps:

### Step 1: Take the natural logarithm of both sides
Given [tex]\( y = e^{x^2} \tan^3(x) \)[/tex], take the natural logarithm (ln) of both sides to simplify the differentiation process:
[tex]\[ \ln(y) = \ln(e^{x^2} \tan^3(x)) \][/tex]

### Step 2: Apply the properties of logarithms
Use the properties of logarithms to separate the product into sums:
[tex]\[ \ln(y) = \ln(e^{x^2}) + \ln(\tan^3(x)) \][/tex]
Since [tex]\(\ln(e^{x^2}) = x^2\)[/tex] and [tex]\(\ln(\tan^3(x)) = 3\ln(\tan(x))\)[/tex], we have:
[tex]\[ \ln(y) = x^2 + 3\ln(\tan(x)) \][/tex]

### Step 3: Differentiate both sides with respect to [tex]\( x \)[/tex]
Differentiate the left side using the chain rule and the right side term-by-term:
[tex]\[ \frac{d}{dx}(\ln(y)) = \frac{d}{dx}(x^2) + \frac{d}{dx}(3\ln(\tan(x))) \][/tex]
[tex]\[ \frac{1}{y} \frac{dy}{dx} = 2x + 3 \frac{d}{dx}(\ln(\tan(x))) \][/tex]

### Step 4: Differentiate [tex]\( \ln(\tan(x)) \)[/tex]
Recall that the derivative of [tex]\( \ln(\tan(x)) \)[/tex] requires using the chain rule:
[tex]\[ \frac{d}{dx}(\ln(\tan(x))) = \frac{1}{\tan(x)} \cdot \sec^2(x) \][/tex]
Simplifying this gives:
[tex]\[ \frac{d}{dx}(\ln(\tan(x))) = \frac{\sec^2(x)}{\tan(x)} = \frac{1}{\sin(x)\cos(x)} \][/tex]

Thus,
[tex]\[ \frac{d}{dx}(3\ln(\tan(x))) = 3 \cdot \frac{\sec^2(x)}{\tan(x)} \][/tex]

### Step 5: Substitute back to the differentiated equation
Our differentiated equation becomes:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = 2x + 3 \frac{\sec^2(x)}{\tan(x)} \][/tex]

### Step 6: Multiply both sides by [tex]\( y \)[/tex]
To isolate [tex]\( \frac{dy}{dx} \)[/tex], multiply both sides of the equation by [tex]\( y \)[/tex]:
[tex]\[ \frac{dy}{dx} = y \left( 2x + 3 \frac{\sec^2(x)}{\tan(x)} \right) \][/tex]

Substitute the value of [tex]\( y = e^{x^2} \tan^3(x) \)[/tex]:
[tex]\[ \frac{dy}{dx} = e^{x^2} \tan^3(x) \left( 2x + 3 \frac{\sec^2(x)}{\tan(x)} \right) \][/tex]

### Step 7: Simplify the expression
Notice that [tex]\( \sec^2(x) = 1 + \tan^2(x) \)[/tex]:
[tex]\[ \frac{dy}{dx} = e^{x^2} \tan^3(x) \left( 2x + 3 \frac{1 + \tan^2(x)}{\tan(x)} \right) \][/tex]
[tex]\[ \frac{dy}{dx} = e^{x^2} \tan^3(x) \left( 2x + 3 \left( \frac{1}{\tan(x)} + \tan(x) \right) \right) \][/tex]
[tex]\[ \frac{dy}{dx} = e^{x^2} \tan^3(x) \left( 2x + 3 \left( \frac{1}{\tan(x)} + \tan(x) \right) \right) \][/tex]
[tex]\[ = e^{x^2} \tan^3(x) \left( 2x + \frac{3}{\tan(x)} + 3 \tan(x) \right) \][/tex]

### Final Simplified Derivative
Thus, the derivative of [tex]\( y = e^{x^2} \tan^3(x) \)[/tex] is:
[tex]\[ \frac{dy}{dx} = e^{x^2} \tan^3(x) \left( 2x + 3 \left( \frac{1}{\tan(x)} + \tan(x) \right) \right) \][/tex]