If `xy^(4)=2x^(2)y^(3)-y^(1//3)-5" at "((dy)/(dx))" at "(2,1)` is

Options

A. `(21)/(47)`
B. `-(21)/(47)`
C. `-(47)/(21)`
D. `(47)/(21)`

Answer

To find \(\frac{dy}{dx}\) at the point \((2, 1)\) for the equation \(xy^4 = 2x^2y^3 - y^{1/3} - 5\), we will use implicit differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate both sides with respect to \(x\)

Starting with the equation:
\[ xy^4 = 2x^2y^3 - y^{1/3} - 5 \]

We differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(xy^4) = \frac{d}{dx}(2x^2y^3) - \frac{d}{dx}(y^{1/3}) - \frac{d}{dx}(5) \]

### Step 2: Apply the product rule on the left side

Using the product rule on the left side:
\[ \frac{d}{dx}(xy^4) = x \frac{d}{dx}(y^4) + y^4 \frac{d}{dx}(x) \]
\[ = x(4y^3 \frac{dy}{dx}) + y^4(1) = 4xy^3 \frac{dy}{dx} + y^4 \]

### Step 3: Differentiate the right side

Now, differentiate the right side:
\[ \frac{d}{dx}(2x^2y^3) = 2x^2 \frac{d}{dx}(y^3) + y^3 \frac{d}{dx}(2x^2) \]
\[ = 2x^2(3y^2 \frac{dy}{dx}) + y^3(4x) = 6x^2y^2 \frac{dy}{dx} + 4xy^3 \]

For the term \(-y^{1/3}\):
\[ \frac{d}{dx}(-y^{1/3}) = -\frac{1}{3}y^{-2/3} \frac{dy}{dx} \]

And the derivative of \(-5\) is \(0\).

### Step 4: Combine the derivatives

Putting it all together, we have:
\[ 4xy^3 \frac{dy}{dx} + y^4 = 6x^2y^2 \frac{dy}{dx} + 4xy^3 - \frac{1}{3}y^{-2/3} \frac{dy}{dx} \]

### Step 5: Collect terms involving \(\frac{dy}{dx}\)

Rearranging gives:
\[ 4xy^3 \frac{dy}{dx} - 6x^2y^2 \frac{dy}{dx} + \frac{1}{3}y^{-2/3} \frac{dy}{dx} = 4xy^3 - y^4 \]

Factoring out \(\frac{dy}{dx}\):
\[ \left(4xy^3 - 6x^2y^2 + \frac{1}{3}y^{-2/3}\right) \frac{dy}{dx} = 4xy^3 - y^4 \]

### Step 6: Solve for \(\frac{dy}{dx}\)

Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4xy^3 - y^4}{4xy^3 - 6x^2y^2 + \frac{1}{3}y^{-2/3}} \]

### Step 7: Substitute the point \((2, 1)\)

Now we substitute \(x = 2\) and \(y = 1\):
\[ \frac{dy}{dx} = \frac{4(2)(1)^3 - (1)^4}{4(2)(1)^3 - 6(2^2)(1^2) + \frac{1}{3}(1)^{-2/3}} \]
\[ = \frac{8 - 1}{8 - 24 + \frac{1}{3}} \]
\[ = \frac{7}{8 - 24 + \frac{1}{3}} = \frac{7}{-16 + \frac{1}{3}} = \frac{7}{-\frac{48}{3} + \frac{1}{3}} = \frac{7}{-\frac{47}{3}} = \frac{7 \cdot 3}{-47} = \frac{21}{-47} \]

Thus, we have:
\[ \frac{dy}{dx} = -\frac{21}{47} \]

### Final Answer
\[ \frac{dy}{dx} \text{ at } (2, 1) = -\frac{21}{47} \]