We will use the quotient rule:
d/dx[f(x)/g(x)]=y'=(f'(x)*g(x)-g'(x)*f(x))/(g(x))^2
But before that however, let's find the derivative of arcsin(3x)
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Let y=arcsin(3x)
Take the sines of both sides
sin(y)=3x
Differentiate both sides W.R.T x
dy/dx*cos(y)=3
Divide cos(y) to both sides
dy/dx=3/cos(y)
We now must rewrite in terms of x
Since , color(blue)(sin(y)=(3x)/1
Then, color(red)(cos(y)=sqrt(1-9x^2)/1=sqrt(1-9x^2)
:.dy/dx=(3)/sqrt(1-9x^2)
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Now finding the derivative:
Let f(x)=arcsin(3x) & g(x)=x
So f'(x)=3/sqrt(1-9x^2) & g'(x)=1
Substituting into the quotient rule we get:
y'=(3/sqrt(1-9x^2)*x-1*arcsin(3x))/(x)^2
Simplify:
y'=((3x)/sqrt(1-9x^2)-arcsin(3x))/x^2
Combine Fractions in the numerator:
color(blue)((3x)/sqrt(1-9x^2)-arcsin(3x)*(sqrt(1-9x^2)/sqrt(1-9x^2))
color(blue)((3x)/sqrt(1-9x^2)-(arcsin(3x)sqrt(1-9x^2))/sqrt(1-9x^2))
y'=((3x-arcsin(3x)sqrt(1-9x^2))/sqrt(1-9x^2))/x^2
Apply the fraction rule for further simplification
y'=(3x-arcsin(3x)sqrt(1-9x^2))/sqrt(1-9x^2)*1/x^2
y'=(3x-arcsin(3x)sqrt(1-9x^2))/(x^2sqrt(1-9x^2))
