What is the second derivative of #f(x)=cos(-x)-sinx#?

Question

Isabella Ackerman · Accepted Answer

#f''(x) = -cosx + sinx# The function can be simplified as #f(x) = cosx - sinx# Differentiating once: #f'(x) = -sinx - cosx# Differentiating again: #f''(x) = -cosx + sinx#

Cameron Alexander · Answer

undefined To find the second derivative of $ f(x) = \cos(-x) - \sin(x) $, we first differentiate the function $ f(x) $ with respect to $ x $ twice.

First derivative:
$$ f'(x) = \frac{d}{dx}[\cos(-x)] - \frac{d}{dx}[\sin(x)] $$

Using the chain rule and derivative of cosine and sine functions:
$$ f'(x) = -(-1)\sin(-x) - \cos(x) $$

$$ f'(x) = \sin(x) - \cos(x) $$

Now, taking the derivative of $ f'(x) $ with respect to $ x $ gives the second derivative:
$$ f''(x) = \frac{d}{dx}[\sin(x) - \cos(x)] $$

$$ f''(x) = \frac{d}{dx}[\sin(x)] - \frac{d}{dx}[\cos(x)] $$

Using the derivative of sine and cosine functions:
$$ f''(x) = \cos(x) + \sin(x) $$

So, the second derivative of $ f(x) = \cos(-x) - \sin(x) $ is $ f''(x) = \cos(x) + \sin(x) $.

Christopher Agresta · Answer

undefined To find the second derivative of $ f(x) = \cos(-x) - \sin(x) $, follow these steps:

1. Find the first derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx} [\cos(-x)] - \frac{d}{dx} [\sin(x)] $$

2. Use the chain rule and derivative of cosine and sine functions:
$$ f'(x) = \sin(-x) - \cos(x) $$

3. Simplify:
$$ f'(x) = -\sin(x) - \cos(x) $$

4. Now, find the second derivative of $ f(x) $:
$$ f''(x) = \frac{d}{dx} [-\sin(x) - \cos(x)] $$

5. Differentiate each term:
$$ f''(x) = -\frac{d}{dx} [\sin(x)] - \frac{d}{dx} [\cos(x)] $$

6. Apply the derivative rules for sine and cosine functions:
$$ f''(x) = -(-\cos(x)) - (-\sin(x)) $$

7. Simplify:
$$ f''(x) = \cos(x) + \sin(x) $$

So, the second derivative of $ f(x) = \cos(-x) - \sin(x) $ is $ f''(x) = \cos(x) + \sin(x) $.