Question

How do you differentiate `y=x(sin(lnx)-cos(lnx))`?

Answer 1

`:.` the answer is `2sin(lnx)`

Explanation:

we have `y=x(sin(lnx)-cos(lnx))` using the product rule to differentiate i.e., `(uv)'=u'v+uv'` on apllying we get
`dy/dx=1(sin(lnx)-cos(lnx))+x(cos(lnx)/x+sin(lnx)/x)rArrdy/dx=sin(lnx)-cos(lnx)+x/x(sin(lnx)+cos(lnx)=sin(lnx)-cos(lnx)+sin(lnx)+cos(lnx)=2sin(lnx)`