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\problemgrouptitle{Differentiation Using All Methods}

\begin{problemblock}[2]{Differentiate the following functions using any appropriate rules.}

\item $\displaystyle f(x)=4x^5+3x^2+x+4$
\answer{$\displaystyle f'(x)=20x^4+6x+1$}

\item $\displaystyle y=x^8+\frac{2}{x^3}-\sqrt{x}+\frac{4}{\sqrt{x}}+10$
\answer{$\displaystyle \frac{dy}{dx}=8x^7-\frac{6}{x^4}-\frac{1}{2}x^{-\frac{1}{2}}-2x^{-\frac{3}{2}}$}

\item $\displaystyle g(t)=\sqrt[3]{t^2}+t^4+\frac{6}{t^2}$
\answer{$\displaystyle g'(t)=\frac{2}{3}t^{-\frac{1}{3}}+4t^3-\frac{12}{t^3}$}

\item $\displaystyle f(x)=\frac{x+5}{\sqrt{x}}$
\answer{$\displaystyle f'(x)=\frac{1}{2}x^{-\frac{1}{2}}-\frac{5}{2}x^{-\frac{3}{2}}$}

\item $\displaystyle g(x)=\left(\sqrt{x}+3x+1\right)\left(x+\pi\right)$
\answer{$\displaystyle g'(x)=\left(\frac{1}{2}x^{-\frac{1}{2}}+3\right)(x+\pi)+\left(\sqrt{x}+3x+1\right)$}

\item $\displaystyle h(y)=\frac{\left(y+4\right)^3}{y+5}$
\answer{$\displaystyle h'(y)=\frac{3(y+4)^2(1+0)(y+5)-(y+4)^3(1+0)}{(y+5)^2}=\frac{(y+4)^2(2y+11)}{(y+5)^2}$}

\item $\displaystyle y=\sqrt[4]{x^3+2x+5}$
\answer{$\displaystyle y'=\frac{1}{4}\left(x^3+2x+5\right)^{-\frac{3}{4}}\left(3x^2+2\right)$}

\item $\displaystyle f(\theta)=\cos 3\theta+\sin^2\theta$
\answer{$\displaystyle f'(\theta)=-3\sin(3\theta)+2\sin\theta\cos\theta$}

\item $\displaystyle g(x)=\tan\left(x^2+1\right)\cos\left(x\right)$
\answer{$g'(x)=2x\sec^2\!\left(x^2+1\right)\cos(x)-\tan\left(x^2+1\right)\sin(x)$}

\item $\displaystyle y=\sqrt[4]{\left(\sin t+5\right)^3}$
\answer{$\displaystyle\frac{dy}{dt}=\frac{3}{4}\left(\sin t+5\right)^{-\frac{1}{4}}\cos t$}

\item $\displaystyle f(x)=\sqrt[3]{\frac{x^4+5x-1}{x^2-3}}$
\answer{$\displaystyle \frac{df}{dx}=\frac{1}{3}\left(\frac{x^4+5x-1}{x^2-3}\right)^{-\frac{2}{3}}\frac{\left(4x^3+5\right)\left(x^2-3\right)-\left(x^4+5x-1\right)\left(2x\right)}{\left(x^2-3\right)^2}\linebreak[0]\newline=\frac{1}{3}\left(\frac{x^4+5x-1}{x^2-3}\right)^{-\frac{2}{3}}\frac{2x^5-12x^3-5x^2+2x-15}{\left(x^2-3\right)^2}$}

\item $\displaystyle x^3y^4+y^2=xy+6$
\answer{$\displaystyle y'=\frac{y-3x^2y^4}{4x^3y^3+2y-x}$}

\end{problemblock}
