How do you differentiate #y=csc^-1(4x^2)#?

Question

Cameron Alexander · Accepted Answer

# dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1)))) # # y=csc^-1(4x^2) <=> cscy=4x^2# Differentiating implicitly we have: # cscy=4x^2 # # -cscycotydy/dx=8x # # -(4x^2)cotydy/dx=8x # # cotydy/dx=-2/x # ..... [1] Now # 1 + cot^2A -= csc^2A # # :. 1 + cot^2y = 16x^4 # # :. cot^2y = 16x^4 - 1 # # :. coty = sqrt(16x^4 - 1) # Substituting into [1] we get: # sqrt(16x^4 - 1)dy/dx=-2/x # # :. dy/dx=(-2/x)/sqrt(16x^4 - 1) # # :. dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1)))) #

Emma Aldridge · Answer

undefined To differentiate $ y = \csc^{-1}(4x^2) $, you can use the chain rule.

The derivative of $ \csc^{-1}(u) $ with respect to $ x $ is $ -\frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx} $.

In this case, $ u = 4x^2 $, so $ \frac{du}{dx} = 8x $.

Substituting these values into the derivative formula, we get:

$$ \frac{dy}{dx} = -\frac{1}{|4x^2|\sqrt{(4x^2)^2 - 1}} \cdot (8x) $$

$$ = -\frac{8x}{|4x^2|\sqrt{16x^4 - 1}} $$

$$ = -\frac{8x}{4x^2\sqrt{16x^4 - 1}} $$

$$ = -\frac{2}{x\sqrt{16x^4 - 1}} $$

Therefore, the derivative of $ y = \csc^{-1}(4x^2) $ with respect to $ x $ is $ -\frac{2}{x\sqrt{16x^4 - 1}} $.