Question

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

Answer 1

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

Explanation:

The derivative of `y` will be the sum of the derivatives of `x^2` and `cos^-1x`. We will find these separately.

DERIVATIVE OF `bb(x^2)`

Since you're expected to find the derivative of `cos^-1x`, it's likely you already know how to differentiate `x^2`. In case you've forgotten, you'll need the power rule.

The power rule states that the derivative of `x^n` is equal to `d/dxx^n=nx^(n-1)`.

So, for `x^2`, we see that the derivative of `x^2` is `d/dxx^2=2x^(2-1)=2x^1=2x`.

DERIVATIVE OF `bb(cos^-1x)`

For this, we will need to do some manipulation. First, let:

`z=cos^-1x`

By the definition of the inverse trig functions (or inverse functions in general) this tells us that

`cos(z)=x`

We now should take the derivative of both sides (with respect to `x`). On the left-hand side, this will require the chain rule since `z` is its own function.

`d/dxcos(z)=d/dxx`

`-sin(z)*(dz)/dx=1`

We then should solve for `dz/dx`, which is the derivative of `cos^-1x`.

`dz/dx=-1/sin(z)`

We can rewrite this in terms of our original function. Remember, `cos(z)=x`. Furthermore, from the Pythagorean identity `sin^2(z)+cos^2(z)=1` we can say that `sin(z)=sqrt(1-cos^2(z))`.

`d/dxcos^-1x=-1/sqrt(1-cos^2(z))`

And since `cos(z)=x`, we can replace `cos^2(z)` with `x^2`:

`d/dxcos^-1x=-1/sqrt(1-x^2)`

PUTTING THEM TOGETHER

We then see that:

`dy/dx=d/dxx^2+d/dxcos^-1x`

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

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