Question

How do you find the derivative of `y = (x ln(x)) / sin(x)`?

Answer 1

`( sinx(1+ lnx) - cosx. xlnx)/sin^2x`

Explanation:

using the`color(blue)(" quotient rule ")`

If f(x) `= g(x)/(h(x))`

then f'(x)`=( h(x)g'(x) - g(x)h'(x) )/(h(x))^2`

now: xlnx is a product of 2 functions and as such will require the`color(blue)(" product rule ")`

If f(x) = g(x)h(x) then f'(x) = g(x)h'(x) + h(x)g'(x)
`color(black)("-----------------------------------------------")`
differentiating the numerator using 'product rule"

`x d/dx(lnx) + lnx d/dx(x) = x.(1/x) + lnx .1= 1 + lnx`

differentiate the denominator: `d/dx(sinx) = cosx`
`color(black)("---------------------------------------------")`

back to the original function using the 'quotient rule'

`dy/dx = (sinx d/dx(xlnx) - xlnx d/dx(sinx))/sin^2x`

'inserting' the derivates gives

`dy/dx =( sinx(1 + lnx) - cosx.xlnx)/sin^2x`