Question

How do you differentiate `y=2cos(x)+6cos^-1(x)`?

Answer 1

Given: `y=2cos(x)+6cos^-1(x)`

Differentiate each term:

`dy/dx=(d(2cos(x)))/dx+(d(6cos^-1(x)))/dx`

Bring the constants outside:

`dy/dx=2(d(cos(x)))/dx+6(d(cos^-1(x)))/dx`

We know that `d/dxcos(x) = -sin(x)`

`dy/dx = -2sin(x) +6(d(cos^-1(x)))/dx`

We know that `d/dxcos^-1(x) = -1/sqrt(1-x^2)`:

`dy/dx = -2sin(x) -6/sqrt(1-x^2)`

Answer 2

`dy/dx=-2sinx-6/sqrt(1-x^2)`

Explanation:

`"using the "color(blue)"standard derivatives"`

`• d/dx(cosx)=-sinx`

`• d/dx(cos^-1x)=-1/(sqrt(1-x^2)`

`y=2cosx+6cos^-1x`

`rArrdy/dx=-2sinx-6/(sqrt(1-x^2)`