How do I find the derivative of #y=sin^-1 (3/(t^2))#?

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Answer 1 · 2018-04-29T05:17:47

#-6/(t^3sqrt(1-(9/t^4))#

Use the chain rule :

#color(purple)(f(g(t))=f^'(g(t))*g^'(t))#

#f=sin^-1# here
#g= 3/t^2# here

#arcsin (sin^-1)= 1/(sqrt(1-(3/t^2)^2))*d/dt[3/t^2]#

For #d/dt[3/t^2]#:

Pull out the 3 because it is a constant so

#3 d/dt [1/t^2]#

#1/t^2# is the same as saying #t^-2# of which it is easy to take the derivative using the power rule :

#d/dt t^-2 = -2t^-3#

so we have #3*(-2t^-3)#

Now you're almost done, simplify:

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

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

#=-6/(t^3sqrt(1-(9/t^4))#