What is the derivative of this function #y=sin^-1(x/3)#?

1 Answer
NickTheTurtle
Apr 6, 2017

#1/(sqrt(9-x^2))#

Explanation:

I am going to assume that the #sin^-1(x)# mean #arcsin(x)# instead of #csc(x)#.

Looking at the differentiation of trigonometric functions, the derivative of #arcsin(x)# is #1/sqrt(1-x^2)#.

Proof:
If #y=arcsin(x)#, then #x=sin(y)#. Differentiating both sides with respect to #x#, we get #1=cos(y)dy/dx#. Then, #dy/dx=1/cos(y)#, which is equal to #1/sqrt(1-sin^2(y))#. But we said that #y=arcsin(x)#, so #dy/dx=1/sqrt(1-x^2)#.

In order to differentiate #arcsin(x/3)#, we use the chain rule: #dy/dx=dy/(du)*(du)/dx#, with #y=arcsin(x/3)# and #u=x/3#.

Then, #d/dx(arcsin(x/3))=d/(d(x/3))(arcsin(x/3))*d/dx(x/3)=1/sqrt(1-(x/3)^2)*1/3=1/(3sqrt(1-1/9x^2))=1/(sqrt(3^2(1-1/9x^2)))=1/(sqrt(9-x^2))#.

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