How do you find the derivative of #sqrt(1/x^3)#?

Question

Ezekiel Alexander · Accepted Answer

Rewrite it. #sqrt(1/(x^3)) = x^(-3/2)#. So the derivative is #-3/2x^(-5/2) = (-3)/(2 sqrt(x^5))#

#sqrt(1/(x^3)) = 1/sqrt(x^3) = 1/x^(3/2) = x^(-3/2)#

Now apply the power rule.

If you really like the quotient rule and algebra , then use

#sqrt(1/(x^3)) = 1/sqrt(x^3) = 1/x^(3/2)# and the quotient rule.

So the derivative is #((0)x^(3/2)-1(3/2x^(1/2)))/(x^(3/2))^2#.

Now simplify algebraically to get # (-3)/(2 sqrt(x^5))#.

If you are a glutton for algebra use the chain rule and the quotient rule:

#d/dx(sqrt(1/x^3)) = d/dx((1/x^3)^(1/2))#

# = 1/2 (1/x^3)^(-1/2) [d/dx(1/x^3)]#

# = 1/2 (1/x^3)^(-1/2) [((0)x^3-1(3x^2))/(x^3)^2]#

Now simplify algebraically.

Luke Alvarez · Answer

undefined To find the derivative of $ \sqrt{\frac{1}{x^3}} $, you can use the chain rule. The chain rule states that if $ f(x) $ and $ g(x) $ are functions, then the derivative of $ f(g(x)) $ with respect to $ x $ is $ f'(g(x)) \cdot g'(x) $.

Let $ f(x) = \sqrt{x} $ and $ g(x) = \frac{1}{x^3} $.

First, find the derivative of $ f(x) = \sqrt{x} $, which is $ f'(x) = \frac{1}{2\sqrt{x}} $.

Next, find the derivative of $ g(x) = \frac{1}{x^3} $, which is $ g'(x) = -3x^{-4} $.

Now, apply the chain rule:

$$
\frac{d}{dx}\left(\sqrt{\frac{1}{x^3}}\right) = f'(g(x)) \cdot g'(x) = \frac{1}{2\sqrt{\frac{1}{x^3}}} \cdot (-3x^{-4})
$$

Simplify:

$$
= \frac{-3}{2x^2\sqrt{x^3}}
$$

So, the derivative of $ \sqrt{\frac{1}{x^3}} $ is $ \frac{-3}{2x^2\sqrt{x^3}} $.