How do you find the derivative of the function: #y = arccosx + x sqrt(1-x^2)#?

Question

Emily Ammerman · Accepted Answer

#dy/dx=(-2x^2)/sqrt(1-x^2# #y=arccos x +xsqrt(1-x^2)# #dy/dx=-1/sqrt(1-x^2)+1*sqrt(1-x^2)+x*1/(2sqrt(1-x^2))*(-2x)# after simplification #dy/dx=(-2x^2)/sqrt(1-x^2#

Scarlett Allison · Answer

undefined To find the derivative of the function $ y = \arccos(x) + x\sqrt{1-x^2} $, you differentiate each term separately using the chain rule and product rule where necessary.

For $\arccos(x)$, its derivative is $-\frac{1}{\sqrt{1-x^2}}$.

For $x\sqrt{1-x^2}$, we have two terms: $x$ and $\sqrt{1-x^2}$. The derivative of $x$ is $1$, and the derivative of $\sqrt{1-x^2}$ is $\frac{-x}{\sqrt{1-x^2}}$.

Therefore, the derivative of the function $y$ is:

$$ y' = -\frac{1}{\sqrt{1-x^2}} + 1 \cdot \sqrt{1-x^2} + x \cdot \frac{-x}{\sqrt{1-x^2}} $$

$$ y' = -\frac{1}{\sqrt{1-x^2}} + \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} $$

$$ y' = -\frac{1}{\sqrt{1-x^2}} - \frac{x^2}{\sqrt{1-x^2}} + \sqrt{1-x^2} $$

$$ y' = -\frac{1 + x^2}{\sqrt{1-x^2}} + \sqrt{1-x^2} $$

Thus, the derivative of $y$ is $y' = -\frac{1 + x^2}{\sqrt{1-x^2}} + \sqrt{1-x^2}$.