How do you find the derivative of #arcsin(x^2/4)#?

Question

Isabella Ackerman · Accepted Answer

# d/dx arcsin(x^2/4) = x/(2sqrt(1-x^4/16)) #

We use the following derivatives: # {: (ul("Function"), qquad ul("Derivative"), qquad ul("Notes")), (f(x), qquad f'(x),), (af(x), qquad af'(x),qquad a " constant"), (x^n, qquad nx^(n-1), qquad n " constant (Power rule)"), (sin^(-1)x, qquad 1/sqrt(1-x^2), ), (f(g(x)), qquad f'(g(x)) \ g'(x),qquad "(Chain rule)" ) :} # So that: # d/dx arcsin(x^2/4) = 1/sqrt(1-(x^2/4)^2) d/dx (x^2/4) # # " " = 1/sqrt(1-x^4/16) * (2x)/4 # # " " = 1/sqrt(1-x^4/16) * x/2 # # " " = x/(2sqrt(1-x^4/16)) #

Aurora Alvarado · Answer

undefined To find the derivative of $ \arcsin \left(\frac{x^2}{4}\right) $, you can use the chain rule. The derivative is:

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

$$ = \frac{1}{\sqrt{1-\left(\frac{x^2}{4}\right)^2}} \cdot \frac{1}{2} \cdot 2x \cdot \frac{1}{4} $$

$$ = \frac{x}{\sqrt{4-x^4}} $$