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

Jan 12, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - \frac{1}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

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

DERIVATIVE OF $\boldsymbol{{x}^{2}}$

Since you're expected to find the derivative of ${\cos}^{-} 1 x$, 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 $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$.

So, for ${x}^{2}$, we see that the derivative of ${x}^{2}$ is $\frac{d}{\mathrm{dx}} {x}^{2} = 2 {x}^{2 - 1} = 2 {x}^{1} = 2 x$.

DERIVATIVE OF $\boldsymbol{{\cos}^{-} 1 x}$

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

$z = {\cos}^{-} 1 x$

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

$\cos \left(z\right) = x$

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.

$\frac{d}{\mathrm{dx}} \cos \left(z\right) = \frac{d}{\mathrm{dx}} x$

$- \sin \left(z\right) \cdot \frac{\mathrm{dz}}{\mathrm{dx}} = 1$

We then should solve for $\frac{\mathrm{dz}}{\mathrm{dx}}$, which is the derivative of ${\cos}^{-} 1 x$.

$\frac{\mathrm{dz}}{\mathrm{dx}} = - \frac{1}{\sin} \left(z\right)$

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

$\frac{d}{\mathrm{dx}} {\cos}^{-} 1 x = - \frac{1}{\sqrt{1 - {\cos}^{2} \left(z\right)}}$

And since $\cos \left(z\right) = x$, we can replace ${\cos}^{2} \left(z\right)$ with ${x}^{2}$:

$\frac{d}{\mathrm{dx}} {\cos}^{-} 1 x = - \frac{1}{\sqrt{1 - {x}^{2}}}$

PUTTING THEM TOGETHER

We then see that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} {x}^{2} + \frac{d}{\mathrm{dx}} {\cos}^{-} 1 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - \frac{1}{\sqrt{1 - {x}^{2}}}$