Higher Order Derivatives — Question 5

Question 5 - Solution

We are given: $$f(x) = \sin(x^2)$$

(a) First three derivatives:

First derivative: $$f'(x) = \frac{d}{dx} \sin(x^2) = \cos(x^2) \cdot 2x = 2x\cos(x^2)$$

Second derivative: Use product rule: $$f''(x) = \frac{d}{dx} \left( 2x \cos(x^2) \right) = 2 \cos(x^2) + 2x \cdot (-\sin(x^2)) \cdot 2x = 2 \cos(x^2) - 4x^2 \sin(x^2)$$

Third derivative: Differentiate $f''(x)$: $$f^{(3)}(x) = \frac{d}{dx} \left( 2 \cos(x^2) - 4x^2 \sin(x^2) \right)$$

First term: $$\frac{d}{dx} [2\cos(x^2)] = -2\sin(x^2) \cdot 2x = -4x \sin(x^2)$$

Second term (product rule): $$\frac{d}{dx} [4x^2 \sin(x^2)] = 8x \sin(x^2) + 4x^2 \cdot \cos(x^2) \cdot 2x = 8x \sin(x^2) + 8x^3 \cos(x^2)$$

So: $$f^{(3)}(x) = -4x \sin(x^2) - \left(8x \sin(x^2) + 8x^3 \cos(x^2)\right) = -12x \sin(x^2) - 8x^3 \cos(x^2)$$

(b) Compute $f''(0)$:

$$f''(x) = 2 \cos(x^2) - 4x^2 \sin(x^2) \Rightarrow f''(0) = 2\cos(0) - 0 = 2$$

(c) Taylor Polynomial $T_2(x)$:

We use: $$T_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2$$

$$f(0) = \sin(0) = 0, \quad f'(0) = 2(0)\cos(0) = 0, \quad f''(0) = 2$$

So: $$T_2(x) = 0 + 0 + \frac{2}{2}x^2 = x^2$$

$$\boxed{T_2(x) = x^2}$$