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Find $f'(x)$ for each function.

Give your answer in the form $\boxed{A}\left(\boxed{\phantom{BB}B\phantom{BB}}\right)^{\boxed{c}}\left(\boxed{\phantom{D}D\phantom{D}}\right)$ without expanding.
Write powers of $x$ in the denominator as negative exponents ($x^{-1}$ or $-3x^{-2}$ and $\color{red}{\text{not } \displaystyle\frac{1}{x} \text{ or } \frac{-3}{x^2}}$)
Write radicals as rational exponents ($x^{\frac{1}{2}}$ or $x^{-\frac{1}{3}}$ and $\color{red}{\text{not } \displaystyle\sqrt{x} \text{ or } \frac{1}{\sqrt[3]{x}}}$)

For example:

$\displaystyle f(x)=\left(\frac{1}{x^2}-\sqrt{x}\right)^5\qquad f'(x)=$

$\boxed {5\left(x^{-2}-x^\frac{1}{2}\right)^4 \left(-2x^{-3}-\frac{1}{2}x^{-\frac{1}{2}}\right)\qquad\qquad\qquad\;}$

$f(x)=\left(4x+5\right)^3 \qquad f'(x)=$

$f(x)=\frac{1}{2}\left(7-3x\right)^5 \qquad f'(x)=$

$f(x)=\left(3x^2-x\right)^2 \qquad f'(x)=$

$f(x) = \left(-2x^3+5x\right)^{-2} \qquad f'(x)=$

$\displaystyle f(x)=\left(\frac{2}{x}+x^3\right)^4\qquad f'(x)=$

$\displaystyle f(x)=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\qquad f'(x)=$

$f(x)=2\left(5x^2-x+7\right)^3\qquad f'(x)=$

$\displaystyle f(x)=-\frac{1}{4}\left(x^3-1\right)^4\qquad f'(x)=$

$f(x)=\left(1-2x\right)^{-5}\qquad f'(x)=$

$\displaystyle f(x)=\frac{4}{\left(3x+1\right)^2}\qquad f'(x)=$

$\displaystyle f(x)=\sqrt{1-2x^3}\qquad f'(x)=$

$f(x)=6\sqrt[3]{4x^2-x}\qquad f'(x)=$

$\displaystyle f(x)=-\frac{4}{\sqrt{x^2+3x+5}}\qquad f'(x)=$

$\displaystyle f(x)=\sqrt{\left(2x^4-x^3\right)^5} \qquad f'(x)=$

$\star\; \displaystyle f(x)=\left(\left(3x^2-1\right)^4 + 5x\right)^3 \qquad f'(x)=$