What is the derivative of #y = arcsin(x^5)#?

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Answer 1 · 2015-08-29T13:37:26

#dy/dx = (5x^4)/sqrt(1-x^10)#

Use the derivative of #arcsin# and the chain rule.

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

When we don't have #x# as the argument of the function, then we need the chain rule:

#d/dx (arcsinu) = 1/sqrt(1-u^2) (du)/dx#

So, we get:

For #y = arcsin(x^5)#,

#dy/dx = 1/sqrt(1-(x^5)^2) d/dx(x^5)#

# = (5x^4)/sqrt(1-x^10)#