Let `y= int_(u(x))^(y(x)) f (t) dt,` let us define `(dy)/(dx) as (dy)/(dx)=v'(x) f (v(x)) - u' (x) f(u(x))` and the equation of the tangent at `(a,b) and y-b=((dy)/(dx))(a,b) (x-a)`.
If ` y=int_(x^(2))^(x^(4)) (In t) dt , "then" lim_(x to 0^(+)) (dy)/(dx)` is equal to

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

To solve the problem, we need to find the limit of \(\frac{dy}{dx}\) as \(x\) approaches \(0^+\) for the function defined by the integral:

\[ y = \int_{x^2}^{x^4} \ln(t) \, dt \]

### Step 1: Identify \(u(x)\) and \(v(x)\)
In our case:
- \(u(x) = x^2\)
- \(v(x) = x^4\)
- \(f(t) = \ln(t)\)

### Step 2: Differentiate \(u(x)\) and \(v(x)\)
Now, we compute the derivatives:
- \(u'(x) = \frac{d}{dx}(x^2) = 2x\)
- \(v'(x) = \frac{d}{dx}(x^4) = 4x^3\)

### Step 3: Apply the Leibniz Rule for Differentiation
Using the Leibniz rule for differentiation under the integral sign, we have:

\[ \frac{dy}{dx} = v'(x) f(v(x)) - u'(x) f(u(x)) \]

Substituting the values we have:

\[ \frac{dy}{dx} = 4x^3 \ln(x^4) - 2x \ln(x^2) \]

### Step 4: Simplify the Expression
Now, we simplify the expression:

\[ \frac{dy}{dx} = 4x^3 (4 \ln(x)) - 2x (2 \ln(x)) \]
\[ = 16x^3 \ln(x) - 4x \ln(x) \]

### Step 5: Factor out common terms
Factoring out \(4x \ln(x)\):

\[ \frac{dy}{dx} = 4x \ln(x) (4x^2 - 1) \]

### Step 6: Find the limit as \(x\) approaches \(0^+\)
Now we need to compute:

\[ \lim_{x \to 0^+} \frac{dy}{dx} = \lim_{x \to 0^+} 4x \ln(x) (4x^2 - 1) \]

### Step 7: Analyze the limit
As \(x \to 0^+\):
- \(4x^2 - 1 \to -1\)
- \(4x \ln(x) \to 0\) (since \(x \ln(x) \to 0\) as \(x \to 0^+\))

Thus, we have:

\[ \lim_{x \to 0^+} 4x \ln(x) (4x^2 - 1) = 0 \cdot (-1) = 0 \]

### Final Answer
Therefore, the limit is:

\[ \lim_{x \to 0^+} \frac{dy}{dx} = 0 \]