If `y=f((3x+4)/(5x+6))andf'(x)=tanx^(2),` then `(dy)/(dx)` is equal to

Options

A. `-2tan((3x+4)/(5x+6))^(2).(1)/((5x+6)^(2))`
B. `f((3tanx^(2)+3)/(5tanx^(2)+6))tanx^(2)`
C. `2xtan((3x-4)/(5x-6))`
D. `tanx^(2)`

Answer

To find \(\frac{dy}{dx}\) for the given function \(y = f\left(\frac{3x + 4}{5x + 6}\right)\) where \(f'(x) = 10x^2\), we will use the chain rule and the quotient rule for differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate using the chain rule
We start by applying the chain rule. The derivative of \(y\) with respect to \(x\) is given by:
\[ \frac{dy}{dx} = f'\left(\frac{3x + 4}{5x + 6}\right) \cdot \frac{d}{dx}\left(\frac{3x + 4}{5x + 6}\right) \]

**Hint:** Remember that the chain rule states that if \(y = f(u)\), then \(\frac{dy}{dx} = f'(u) \cdot \frac{du}{dx}\).

### Step 2: Differentiate the inner function using the quotient rule
Now, we need to differentiate \(\frac{3x + 4}{5x + 6}\) using the quotient rule. The quotient rule states that if \(y = \frac{u}{v}\), then:
\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
Here, \(u = 3x + 4\) and \(v = 5x + 6\).

Calculating the derivatives:
- \(\frac{du}{dx} = 3\)
- \(\frac{dv}{dx} = 5\)

Now applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{3x + 4}{5x + 6}\right) = \frac{(5x + 6)(3) - (3x + 4)(5)}{(5x + 6)^2} \]

**Hint:** Make sure to apply the quotient rule correctly by substituting the derivatives of \(u\) and \(v\).

### Step 3: Simplify the expression
Now, simplify the numerator:
\[ = \frac{(15x + 18) - (15x + 20)}{(5x + 6)^2} = \frac{15x + 18 - 15x - 20}{(5x + 6)^2} = \frac{-2}{(5x + 6)^2} \]

**Hint:** Combine like terms carefully to avoid mistakes in simplification.

### Step 4: Substitute back into the derivative expression
Now substitute this back into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = f'\left(\frac{3x + 4}{5x + 6}\right) \cdot \frac{-2}{(5x + 6)^2} \]

### Step 5: Substitute \(f'\) using the given information
We know \(f'(x) = 10x^2\). Thus:
\[ f'\left(\frac{3x + 4}{5x + 6}\right) = 10\left(\frac{3x + 4}{5x + 6}\right)^2 \]

### Step 6: Final expression for \(\frac{dy}{dx}\)
Substituting this into our derivative gives:
\[ \frac{dy}{dx} = 10\left(\frac{3x + 4}{5x + 6}\right)^2 \cdot \frac{-2}{(5x + 6)^2} \]
\[ = \frac{-20\left(3x + 4\right)^2}{(5x + 6)^4} \]

### Final Answer
Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{-20(3x + 4)^2}{(5x + 6)^4} \]