What is the derivative of #f(x) = arcsin(2x^3 - 1)#?

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Answer 1 · 2018-04-13T08:43:57

#(df)/(dx)=(3x^2)/(xsqrt(x(1-x^3))#

We can use here chain formula. As differential of #arcsinx# is #1/sqrt(1-x^2)#

derivative of #f(x)=arcsin(2x^3-1)# is

#(df)/(dx)=1/sqrt(1-(2x^3-1)^2)*d/(dx)(2x^3-1)#

= #1/sqrt(1-4x^6+4x^3-1)*6x^2#

= #(3x^2)/(xsqrt(x(1-x^3))#

Answer 2 · 2018-04-13T12:32:13

#(3x^2)/sqrt(x^3(1-x^3)#

We use the chain rule, which states that,

#dy/dx=dy/(du)*(du)/dx#

Let #u=2x^3-1,:.(du)/dx=6x^2#.

Then #y=arcsinu,=>dy/(du)=1/sqrt(1-u^2)#.

Combining together, we get,

#dy/dx=1/sqrt(1-u^2)*6x^2#

#=(6x^2)/sqrt(1-u^2)#

Final step is to substitute back #u=2x^3-1#, and we get,

#=(6x^2)/sqrt(1-(2x^3-1)^2)#

#=(3x^2)/sqrt(x^3(1-x^3)#