Question

Question #5f32d

Answer 1

`-cotxcscx`

Explanation:

We have `d/dx1/sinx`

We can apply the quotient rule, that is, `(f/g)'`, where Newton's notation is used and `f` and `g` are functions, is equal to `(f'g-fg')/g^2`.

Here, `f=1` and `g=sinx`. We can input:

`(d/dx(1)*sinx-1*d/dx(sinx))/sin^2x`

We must find `f'` and `g'`.

`d/dx1=0`, so the above equation reduces to:

`(-d/dxsinx)/sin^2x`

Since `d/dxsinx=cosx`, we can input:

`-cosx/sin^2x`

Or:

`-cosx/sinx*1/sinx`

Since `cosx/sinx=cotx` and `1/sinx=cscx`, this becomes:

`-cotxcscx`

Answer 2

`-cotxcscx`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x)xxg'(x)`

`"here "y=1/sinx=(sinx)^-1`

`rArrdy/dx=-(sinx)^-2xxd/dx(sinx)`

`color(white)(rArrdy/dx)=-cosx(sinx)^-2`

`color(white)(rArrdy/dx)=-cosx/(sin^2x`

`color(white)(rArrdy/dx)=-cosx/sinx xx1/sinx`

`color(white)(rArrdy/dx)=-cotxcscx`