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

Answer 1

#-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))#
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Answer 2

To find the derivative of (y = \sin^{-1}\left(\frac{3}{t^2}\right)), use the chain rule. The derivative is (-\frac{6}{t^3\sqrt{1-\left(\frac{3}{t^2}\right)^2}}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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