what is the derivative of y=(lnx^2)^2?

Answer 1

2lnx^2*1/x^2*2x
4lnx^2/x

Answer 2

You need to apply the chain rule twice:
( ... )^2 is the outermost function.
ln( ... ) is next,
and x^2 is innermost.
The derivative of the outermost function is 2*( ln(x^2) )
You multiply this by the derivative of ln(x^2), which is 1/(x^2) multiplied by the derivative of x^2 which is 2x.
So we have 2*( ln(x^2) ) * 1/( x^2 ) * 2x.

Of course this will simplify to:
4*(ln(x^2))/x

Answer 3

Mostly it's d/dx(f(x))^n = n(f(x))^(n-1)*d/dx(f(x)) used over and over.
Going through it stepwise:
d/dx(ln(x²))²
2(ln(x²)) * d/dx(ln(x²))
2(ln(x²)) * (1/x²) * d/dx x²
2(ln(x²)) * (1/x²) * 2x
4ln(x²)/x

Hope that helps ☺

Doug