What is the second derivative of #f(x)=sin(1/x^2)
#?

Question

What is the second derivative of #f(x)=sin(1/x^2)
 #?

Ezra Adams · Accepted Answer

#f'(x)=cos(1/x)*-1/x^2#
#f=cos(1/x), g= -1/x^2-> f'=-sin(1/x)(-1/x^2), g'=2/x^3 #
#f''(x)=fg'+gf'=cos(1/x)(2/x^3)+(-1/x^2)sin(1/x)(1/x^2)=2/x^3cos(1/x^2)+(-1/x^4)sin(1/x)# Find the first derivative using chain rule. To find the second derivative use the product rule. So separate the functions into f and g then find f' and g' and then apply the product rule.

Elijah Abram · Answer

undefined To find the second derivative of $ f(x) = \sin\left(\frac{1}{x^2}\right) $, you need to first find the first derivative and then differentiate it again.

1. Find the first derivative:
$$ f'(x) = \frac{d}{dx}\left[\sin\left(\frac{1}{x^2}\right)\right] $$

Using the chain rule, the derivative of $\sin(u)$ is $\cos(u)$ multiplied by the derivative of $u$:
$$ f'(x) = \cos\left(\frac{1}{x^2}\right) \cdot \frac{d}{dx}\left(\frac{1}{x^2}\right) $$

Differentiate $\frac{1}{x^2}$:
$$ \frac{d}{dx}\left(\frac{1}{x^2}\right) = -2x^{-3} $$

$$ f'(x) = \cos\left(\frac{1}{x^2}\right) \cdot (-2x^{-3}) $$
$$ f'(x) = -2x^{-3} \cos\left(\frac{1}{x^2}\right) $$

2. Now, find the second derivative by differentiating $f'(x)$ with respect to $x$:
$$ f''(x) = \frac{d}{dx}\left[-2x^{-3} \cos\left(\frac{1}{x^2}\right)\right] $$

Using the product rule and the chain rule:
$$ f''(x) = -2 \cdot (-3)x^{-4} \cos\left(\frac{1}{x^2}\right) + (-2x^{-3}) \cdot \frac{d}{dx}\left(\cos\left(\frac{1}{x^2}\right)\right) $$

$$ f''(x) = 6x^{-4} \cos\left(\frac{1}{x^2}\right) + 2x^{-3} \sin\left(\frac{1}{x^2}\right) \cdot \frac{d}{dx}\left(\frac{1}{x^2}\right) $$

Differentiate $\frac{1}{x^2}$ to get:
$$ \frac{d}{dx}\left(\frac{1}{x^2}\right) = -2x^{-3} $$

$$ f''(x) = 6x^{-4} \cos\left(\frac{1}{x^2}\right) + 2x^{-3} \sin\left(\frac{1}{x^2}\right) \cdot (-2x^{-3}) $$

$$ f''(x) = 6x^{-4} \cos\left(\frac{1}{x^2}\right) - 4x^{-6} \sin\left(\frac{1}{x^2}\right) $$

Therefore, the second derivative of $ f(x) = \sin\left(\frac{1}{x^2}\right) $ is $ f''(x) = 6x^{-4} \cos\left(\frac{1}{x^2}\right) - 4x^{-6} \sin\left(\frac{1}{x^2}\right) $.

What is the second derivative of #f(x)=sin(1/x^2) #?