Question

How do you differentiate `f(x)=sin^2(lnx)xcos^2(x^2)` using the chain rule?

Answer 1

#=-4x^2sin^2(lnx)sin(x^2)cos(x^2)+sin^2(lnx)cos^2(x^2)+xsin(2lnx)cos^2(x^2)

Explanation:

`=[sin^2(lnx)][d/dx(xcos^2(x^2))]+[d/dx(sin^2(lnx))][xcos^2(x^2)]`
`=[sin^2(lnx)][(x)(d/dx(cos^2(x^2)))+(1)(cos^2(x^2))]+[(2sin(lnx)cos(lnx))/x][cos^2(x^2)]`
`=[sin^2(lnx)][(x)(-2sin(x^2)cos(x^2)(2x))+cos^2(x^2)]+[sin(2lnx)cos^2(x^2)]`
`=-4x^2sin^2(lnx)sin(x^2)cos(x^2)+sin^2(lnx)cos^2(x^2)+xsin(2lnx)cos^2(x^2)`

The chain rule is very messy here but this should guide you:
For `sin^2(lnx)`, differentiate the quadratic term, triginometric term, `ln` term and lastly the `x` term.

For `xsin^2(x^2)` Use Product Rule to separate `x`. Differentiate the quadratic term, then the trigonometric term then the inside `(x^2)` term.
`d/dx[cos^2(x^2)]->2cos(x^2)->-2cos(x^2)sin(x^2)->-4xsin(x^2)cos(x^2)`