If `3f(cosx)+2f(sinx)=5x`, then `f^(')(cosx)` is equal to (where `f^(')` denotes derivative with respect to`x`)

Options

A. `-5/(cosx)`
B. `5/(cosx)`
C. `-5/(sinx)`
D. `5/(sinx)`

Explanation

To solve the problem, we start with the equation given:

\[ 3f(\cos x) + 2f(\sin x) = 5x \]

### Step 1: Substitute \( x \) with \( \frac{\pi}{2} - x \)

We substitute \( x \) with \( \frac{\pi}{2} - x \) in the original equation:

\[ 3f(\cos(\frac{\pi}{2} - x)) + 2f(\sin(\frac{\pi}{2} - x)) = 5\left(\frac{\pi}{2} - x\right) \]

Using the trigonometric identities \( \cos(\frac{\pi}{2} - x) = \sin x \) and \( \sin(\frac{\pi}{2} - x) = \cos x \), we can rewrite the equation as:

\[ 3f(\sin x) + 2f(\cos x) = \frac{5\pi}{2} - 5x \]

### Step 2: Label the equations

Let’s label the original equation as Equation (1):

\[ 3f(\cos x) + 2f(\sin x) = 5x \quad \text{(1)} \]

And the substituted equation as Equation (2):

\[ 3f(\sin x) + 2f(\cos x) = \frac{5\pi}{2} - 5x \quad \text{(2)} \]

### Step 3: Multiply Equation (1) and Equation (2)

Multiply Equation (1) by 3:

\[ 9f(\cos x) + 6f(\sin x) = 15x \quad \text{(3)} \]

Multiply Equation (2) by 2:

\[ 6f(\sin x) + 4f(\cos x) = 5\pi - 10x \quad \text{(4)} \]

### Step 4: Subtract Equation (4) from Equation (3)

Now, we will subtract Equation (4) from Equation (3):

\[ (9f(\cos x) + 6f(\sin x)) - (6f(\sin x) + 4f(\cos x)) = 15x - (5\pi - 10x) \]

This simplifies to:

\[ (9f(\cos x) - 4f(\cos x)) = 15x + 10x - 5\pi \]

\[ 5f(\cos x) = 25x - 5\pi \]

### Step 5: Solve for \( f(\cos x) \)

Dividing both sides by 5 gives:

\[ f(\cos x) = 5x - \pi \]

### Step 6: Differentiate \( f(t) \)

Now, we will differentiate \( f(t) \) with respect to \( t \):

Let \( t = \cos x \), then \( f(t) = 5\cos^{-1}(t) - \pi \).

Differentiating \( f(t) \):

\[ f'(t) = \frac{d}{dt}(5\cos^{-1}(t)) = \frac{-5}{\sqrt{1 - t^2}} \]

### Step 7: Substitute back \( t = \cos x \)

Now substituting back \( t = \cos x \):

\[ f'(\cos x) = \frac{-5}{\sqrt{1 - \cos^2 x}} = \frac{-5}{\sin x} \]

### Final Answer

Thus, we find that:

\[ f'(\cos x) = \frac{-5}{\sin x} \]