Find the Derivative and simplify (write w/o negative exponents and factor if possible) y=cos^3(x^2+2) I am currently practicing on an old college calculus final. Would like to know if my work is correct. 
Find the derivative for f(x) = x^1/2 + (3/x^2)-2 and y=cos^3(x^2+2) . Thanks for your help.


 A: You got one part right, but there are a few mistakes. For example, the function $\frac{3}{x^2}-2$ is equivalently written as $3x^{-2}-2$. By the power rule, the derivative would be $$(-2)3x^{-2-1} = -6x^{-3} = \frac{-6}{x^3}$$ Your derivative of $x^{1/2}$ is correct. 
The derivative of $\cos^3(x^3+2)$ is wrong for a few reasons. First, you are missing one "layer" in your chain rule. Second, it is the cosine argument being raised to the $3$rd power, not the quantity $(x^3+2)$ inside the argument of cosine. You can think of it as $$\left[\cos(x^3+2) \right]^3 = \cos(x^3+2)\cdot \cos(x^3+2) \cdot \cos(x^3+2)$$ By the power rule, you should get $$\frac{d}{dx}\left[\cos(x^3+2) \right]^3  = 3\left[\cos(x^3+2) \right]^2 \cdot \left[\frac{d}{dx} \cos(x^3+2)\right]$$ In differentiating $\frac{d}{dx}\cos(x^3+2)$ you should get $$-\sin(x^3+2) \cdot \left[\frac{d}{dx}(x^3+2)\right] = -\sin(x^3+2)\cdot 3x^2$$ So all together, $$\frac{d}{dx}\left[\cos(x^3+2) \right]^3 = -9x^2\left[\cos(x^3+2) \right]^2\sin(x^3+2)$$
A: Hint:
$$
\dfrac{d}{dx}\dfrac{1}{x^2}=\dfrac{d}{dx}x^{-2}=-2x^{-3}
$$
In the second you have an initial mistake:
$$
\dfrac{d}{dx}\cos^3(x^3+2)=\dfrac{d}{dx}\left[ \cos(x^3+2)\right]^3=3\left[ \cos(x^3+2)\right]^2\dfrac{d}{dx}\cos(x^3+2)=....
$$
