Question

Question #217eb

Answer 1

`f'(x) = (sin^2x + sinx + xcosx)/(1 + sinx)^2`

Explanation:

First of all, the function can be simplified.

Call the function `f(x)`.

`f(x) = (xtanx)/(1/cosx + sinx/cosx)`

`f(x) = (xtanx)/((1 + sinx)/cosx)`

`f(x) = (xtanx(cosx))/(1 + sinx)`

`f(x) = (xsinx/cosx(cosx))/(1 + sinx)`

`f(x) = (xsinx)/(1 + sinx)`

We now differentiate this using the quotient rule. The quotient rule states that for a function `color(red)(f(x) = (g(x))/(h(x))`, the derivative is given by `color(red)(f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2`. We know the derivative of the numerator is `sinx + xcosx` (by the product rule), and that the derivative of the denominator is `cosx`.

Hence:

`f'(x) = ((sinx + xcosx)(1 + sinx) - xsinxcosx)/(1 + sinx)^2`

`f'(x) = (sinx + xcosx+ sin^2x + xcosxsinx - xcosxsinx)/(1 + sinx)^2`

`f'(x) = (sin^2x + sinx + xcosx)/(1 + sinx)^2`

Hopefully this helps!