How do you differentiate # f(x) =arccos(2x + 1) #?

Question

How do you differentiate # f(x) =arccos(2x + 1)   #?

Emma Aldridge · Accepted Answer

# f'(x)=-1/sqrt(-x^2-x).# Recall that, #d/dt arc cos t=-1/sqrt(1-t^2).# Therefore, #f(x)-arc cos(2x+1)# # rArr f'(x)=-1/sqrt{1-(2x+1)^2}*d/dx(2x+1),...[because," The Chain Rule]."# # rArr f'(x)=-2/sqrt(-4x^2-4x)=-1/sqrt(-x^2-x).#

Isaiah Allen · Answer

undefined To differentiate $ f(x) = \arccos(2x + 1) $, you can use the chain rule, which states that if $ u $ is a differentiable function of $ x $, and $ y $ is a differentiable function of $ u $, then $ y $ is a differentiable function of $ x $ and its derivative is given by $ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $.

First, identify $ u $ as the inner function, which is $ u = 2x + 1 $. Then, differentiate $ u $ with respect to $ x $, yielding $ \frac{du}{dx} = 2 $.

Next, differentiate $ y = \arccos(u) $ with respect to $ u $, resulting in $ \frac{dy}{du} = -\frac{1}{\sqrt{1-u^2}} $.

Apply the chain rule formula to find the derivative of $ f(x) $:
$$ \frac{df}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{\sqrt{1-(2x + 1)^2}} \cdot 2 $$

Finally, simplify the expression to get the derivative:
$$ \frac{df}{dx} = -\frac{2}{\sqrt{1-(2x + 1)^2}} $$