If `f(x)=x^(3)tanx+3x^(4)sinx`, find its derivative at x = 0 and `x = pi/4`.

Answer

To find the derivative of the function \( f(x) = x^3 \tan x + 3x^4 \sin x \) at \( x = 0 \) and \( x = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Differentiate the function \( f(x) \)

We will use the product rule for differentiation, which states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then:

\[ (uv)' = u'v + uv' \]

For our function, we can identify:

1. \( u_1 = x^3 \) and \( v_1 = \tan x \)
2. \( u_2 = 3x^4 \) and \( v_2 = \sin x \)

Now, we will differentiate each part.

#### Part 1: Differentiate \( x^3 \tan x \)

Using the product rule:

\[ \frac{d}{dx}(x^3 \tan x) = (3x^2)(\tan x) + (x^3)(\sec^2 x) \]

#### Part 2: Differentiate \( 3x^4 \sin x \)

Using the product rule:

\[ \frac{d}{dx}(3x^4 \sin x) = (12x^3)(\sin x) + (3x^4)(\cos x) \]

### Step 2: Combine the derivatives

Now, we combine both parts to find \( f'(x) \):

\[ f'(x) = 3x^2 \tan x + x^3 \sec^2 x + 12x^3 \sin x + 3x^4 \cos x \]

### Step 3: Evaluate \( f'(0) \)

Now we will evaluate the derivative at \( x = 0 \):

\[ f'(0) = 3(0)^2 \tan(0) + (0)^3 \sec^2(0) + 12(0)^3 \sin(0) + 3(0)^4 \cos(0) \]

Since all terms contain a factor of \( 0 \), we have:

\[ f'(0) = 0 + 0 + 0 + 0 = 0 \]

### Step 4: Evaluate \( f'\left(\frac{\pi}{4}\right) \)

Now we will evaluate the derivative at \( x = \frac{\pi}{4} \):

1. Calculate \( \tan\left(\frac{\pi}{4}\right) = 1 \)
2. Calculate \( \sec^2\left(\frac{\pi}{4}\right) = 2 \)
3. Calculate \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
4. Calculate \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)

Now substitute these values into \( f'\left(\frac{\pi}{4}\right) \):

\[ f'\left(\frac{\pi}{4}\right) = 3\left(\frac{\pi}{4}\right)^2(1) + \left(\frac{\pi}{4}\right)^3(2) + 12\left(\frac{\pi}{4}\right)^3\left(\frac{\sqrt{2}}{2}\right) + 3\left(\frac{\pi}{4}\right)^4\left(\frac{\sqrt{2}}{2}\right) \]

Calculating each term:

1. \( 3\left(\frac{\pi}{4}\right)^2 = \frac{3\pi^2}{16} \)
2. \( \left(\frac{\pi}{4}\right)^3(2) = \frac{2\pi^3}{64} = \frac{\pi^3}{32} \)
3. \( 12\left(\frac{\pi}{4}\right)^3\left(\frac{\sqrt{2}}{2}\right) = 12 \cdot \frac{\pi^3}{64} \cdot \frac{\sqrt{2}}{2} = \frac{12\pi^3\sqrt{2}}{128} = \frac{3\pi^3\sqrt{2}}{32} \)
4. \( 3\left(\frac{\pi}{4}\right)^4\left(\frac{\sqrt{2}}{2}\right) = 3 \cdot \frac{\pi^4}{256} \cdot \frac{\sqrt{2}}{2} = \frac{3\pi^4\sqrt{2}}{512} \)

### Step 5: Combine the results

Now combine all the terms:

\[ f'\left(\frac{\pi}{4}\right) = \frac{3\pi^2}{16} + \frac{\pi^3}{32} + \frac{3\pi^3\sqrt{2}}{32} + \frac{3\pi^4\sqrt{2}}{512} \]

This gives us the final value of the derivative at \( x = \frac{\pi}{4} \).

### Final Answers:

- \( f'(0) = 0 \)
- \( f'\left(\frac{\pi}{4}\right) = \frac{3\pi^2}{16} + \frac{\pi^3}{32} + \frac{3\pi^3\sqrt{2}}{32} + \frac{3\pi^4\sqrt{2}}{512} \)