What is the derivative of #f(x)=(cos^-1(x))/x# ?

Question

Ezekiel Alexander · Accepted Answer

#f'(x)=-1/(xsqrt(1-x^2))-(cos^-1x)/x^2# Using Quotient Rule, which is  #y=f(x)/g(x)#, then #y'=(f'(x)g(x)−f(x)g'(x))/(g(x))^2# Applying this for given problem, which is #f(x)=(cos^-1x)/x# #f'(x)=((cos^-1x)'(x)-(cos^-1x)(x)')/x^2# #f'(x)=(-1/sqrt(1-x^2)*x-cos^-1x)/x^2# #f'(x)=-1/(xsqrt(1-x^2))-(cos^-1x)/x^2#, where #-1#<#x#<#1#

Isabella Ackerman · Answer

undefined To find the derivative of $ f(x) = \frac{\cos^{-1}(x)}{x} $, we can use the quotient rule. The quotient rule states that if $ u(x) $ and $ v(x) $ are differentiable functions, then the derivative of $ \frac{u(x)}{v(x)} $ is given by:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

In this case, $ u(x) = \cos^{-1}(x) $ and $ v(x) = x $. To differentiate $ \cos^{-1}(x) $, we can use the chain rule, which states that if $ f(x) $ is differentiable and $ g(x) $ is differentiable at $ f(x) $, then the derivative of $ g(f(x)) $ is $ g'(f(x)) \cdot f'(x) $.

Therefore, the derivative of $ \cos^{-1}(x) $ is $ -\frac{1}{\sqrt{1-x^2}} $ (you can verify this using the chain rule).

Now, applying the quotient rule, we have:

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

Simplifying, we get:

$$ f'(x) = \frac{-x}{x\sqrt{1-x^2}} - \frac{\cos^{-1}(x)}{x^2} $$

$$ f'(x) = -\frac{1}{\sqrt{1-x^2}} - \frac{\cos^{-1}(x)}{x^2} $$

So, the derivative of $ f(x) = \frac{\cos^{-1}(x)}{x} $ is $ -\frac{1}{\sqrt{1-x^2}} - \frac{\cos^{-1}(x)}{x^2} $.