1. Take the derivative like a regular derivative; however, when we take the derivative of the dependent variable, we'll multiply it by \(y'\). \[\frac{d}{dx}\left[x^{2}+y^{2}\right]=\frac{d}{dx}\left[25\right]\] \[2x+2y\cdot y'=0\]
2. Next, solve for \(y'\). \[2y\cdot y'=-2x\] \[y'=-\frac{2x}{2y}=-\frac{x}{y}\]

1. Take the derivative \[\frac{d}{dx}\left[e^{-x}+e^{y}\right]=\frac{d}{dx}\left[4x^{2}\right]\] \[-e^{-x}+e^{y}\cdot y'=8x\]
2. Solve for \(y'\) \[e^{y}\cdot y'=8x-e^{-x}\] \[y'=\frac{8x-e^{-x}}{e^{y}}\]

1. Take the derivative \[\frac{1}{y}\cdot\frac{dy}{dx}=\frac{5x^{4}}{2\sqrt{x^{5}}}\] \[\frac{dy}{dx}=\frac{5x^{4}y}{2\sqrt{x^{5}}}\]
2. Take the derivative of the above expression \[\frac{d}{dx}\left[\frac{dy}{dx}\right]=\frac{d}{dx}\left[\frac{5x^{4}y}{2\sqrt{x^{5}}}\right]\] \[\frac{d^{2}y}{dx^{2}}=\frac{\left(\frac{d}{dx}\left[5x^{4}y\right]\right)\left(2\sqrt{x^{5}}\right)-\left(\frac{d}{dx}\left[2\sqrt{x^{5}}\right]\right)\left(5x^{4}y\right)}{\left(2\sqrt{x^{5}}\right)^{2}}=\] \[\frac{\left(\frac{d}{dx}\left[5x^{4}\right]\left(y\right)+\left(5x^{4}\right)\frac{d}{dx}\left[y\right]\right)\left(2\sqrt{x^{5}}\right)-\left(5x^{4}y\cdot\frac{2\cdot5x^{4}}{2\sqrt{x^{5}}}\right)}{4x^{5}}=\] \[\frac{\left(20x^{3}y+\left(\frac{dy}{dx}\cdot5x^{4}\right)\right)\left(2\sqrt{x^{5}}\right)-\frac{25x^{8}y}{\sqrt{x^{5}}}}{4x^{5}}=\] Substitute the first derivative in for \(\frac{dy}{dx}\) and simplify. \[\frac{\left(20x^{3}y+\left(\frac{5x^{4}y}{2\sqrt{x^{5}}}\cdot5x^{4}\right)\right)\left(2\sqrt{x^{5}}\right)-\frac{25x^{8}y}{\sqrt{x^{5}}}}{4x^{5}}=\] \[\frac{\left(20x^{3}y+\frac{25x^{8}y}{2\sqrt{x^{5}}}\right)\left(2\sqrt{x^{5}}\right)-\frac{25x^{8}y}{\sqrt{x^{5}}}}{4x^{5}}=\] \[\frac{40\sqrt{x^{5}}x^{3}y+25x^{8}y-\frac{25x^{8}y}{\sqrt{x^{5}}}}{4x^{5}}=\frac{25x^{8}y+\left(40\sqrt{x^{5}}x^{3}y\right)\left(\frac{\sqrt{x^{5}}}{\sqrt{x^{5}}}\right)-\frac{25x^{8}y}{\sqrt{x^{5}}}}{4x^{5}}=\] \[\frac{25x^{8}y+\frac{15x^{8}y}{\sqrt{x^{5}}}}{4x^{5}}=\frac{25x^{3}y+\frac{15x^{3}y}{\sqrt{x^{5}}}}{4}=\frac{25x^{3}y+15\left(x^{3}\right)\left(x^{-\frac{5}{2}}\right)y}{4}=\] \[\frac{25x^{3}y+15\left(x^{\frac{1}{2}}\right)y}{4}=\frac{25x^{3}y+15y\sqrt{x}}{4}\] \[\therefore\] \[y''=\frac{25x^{3}y+15y\sqrt{x}}{4}\]

1. Take the derivative \[\frac{d}{dx}\left[\frac{\left(y-3\right)^{2}}{6}-\frac{\left(x-4\right)^{2}}{8}\right]=\frac{d}{dx}\left[1\right]\] \[\left(\frac{2\left(y-3\right)}{6}\cdot\frac{dy}{dx}\right)-\frac{2\left(x-4\right)}{8}=0\] \[\left(\frac{y-3}{3}\cdot\frac{dy}{dx}\right)-\frac{x-4}{4}=0\]
2. Solve for \(\frac{dy}{dx}\) \[\frac{y-3}{3}\cdot\frac{dy}{dx}=\frac{x-4}{4}\] \[\frac{dy}{dx}=\frac{x-4}{4}\cdot\frac{3}{y-3}=\frac{3\left(x-4\right)}{4\left(y-3\right)}\]
3. Evaluate the derivative at the point \[\left.\frac{dy}{dx}\right|_{(6,0)} \frac{3\left(x-4\right)}{4\left(y-3\right)}=\frac{3\left(6-4\right)}{4\left(0-3\right)}=\frac{6}{-12}=-\frac{1}{2}\]

We need to find where the derivative does not exist.

1. Take the derivative and solve for \(y'\) \[y'=\frac{1}{2y}\]
2. The above expression does not exist when \(y=0\); therefore, we need to solve the original equation for \(y\) and determine where it is equal to \(0\) or is discontinuous. \[y=\pm\sqrt{x-3}\] The above expression is square root function is equal to \(0\) when \(x=3\) and is undefined when \(x

Since one function is to the power of another function, we need take the natural log of both sides of the equation. Then we can find the derivative by using implicit differentiation.

\[y=x^{4\sqrt{x}}\]

\[\ln\left(y\right)=\ln\left(x^{4\sqrt{x}}\right)\]

\[\ln\left(y\right)=4\sqrt{x}\ln\left(x\right)\]

\[\frac{d}{dx}\left[\ln\left(y\right)\right]=\frac{d}{dx}\left[4\sqrt{x}\ln\left(x\right)\right]\]

\[\frac{1}{y}\cdot\frac{dy}{dx}=\ln\left(x\right)\frac{d}{dx}\left[4\sqrt{x}\right]+4\sqrt{x}\frac{d}{dx}\left[\ln\left(x\right)\right]\]

\[\frac{1}{y}\cdot\frac{dy}{dx}=\ln\left(x\right)\left(\frac{4}{2\sqrt{x}}\right)+4\sqrt{x}\left(\frac{1}{x}\right)\]

\[\frac{1}{y}\cdot\frac{dy}{dx}=\frac{2\ln\left(x\right)}{\sqrt{x}}+\frac{4\sqrt{x}}{x}=\frac{2\ln\left(x\right)\sqrt{x}}{\sqrt{x}\sqrt{x}}+\frac{4\sqrt{x}}{x}=\]

\[\frac{2\ln\left(x\right)\sqrt{x}}{x}+\frac{4\sqrt{x}}{x}=\frac{2\ln\left(x\right)\sqrt{x}+4\sqrt{x}}{x}\]

\[\frac{1}{y}\cdot\frac{dy}{dx}=\frac{2\ln\left(x\right)\sqrt{x}+4\sqrt{x}}{x}\]

\[\frac{dy}{dx}=\frac{y\left(2\ln\left(x\right)\sqrt{x}+4\sqrt{x}\right)}{x}\]

Since we aren't asked to compute any higher order derivatives, we can substitute in the original equation for \(y\)

\[\therefore\]

\[\frac{dy}{dx}=\frac{\left(x^{4\sqrt{x}}\right)\left(2\ln\left(x\right)\sqrt{x}+4\sqrt{x}\right)}{x}\]

1. Find the value of the function at \(x=\frac{\pi}{2}\) \[f\left(\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}}=1^{\frac{\pi}{2}}=1\]
2. Take the natural log of each side \[\ln\left(y\right)=\ln\left(\sin\left(x\right)^{x}\right)\] \[\ln\left(y\right)=x\ln\left(\sin\left(x\right)\right)\]
3. Find the derivative using implicit differentiation \[\frac{d}{dx}\left[\ln\left(y\right)\right]=\frac{d}{dx}\left[x\ln\left(\sin\left(x\right)\right)\right]\] \[\frac{1}{y}\cdot y'=\left(\ln\left(\sin\left(x\right)\right)\right)\frac{d}{dx}\left[x\right]+x\frac{d}{dx}\left[\ln\left(\sin\left(x\right)\right)\right]\] \[\frac{1}{y}\cdot y'=\ln\left(\sin\left(x\right)\right)+\left(x\right)\frac{1}{\sin\left(x\right)}\left(\cos\left(x\right)\right)=\ln\left(\sin\left(x\right)\right)+\frac{x\cos\left(x\right)}{\sin\left(x\right)}\] \[y'=y\left(\ln\left(\sin\left(x\right)\right)+\frac{x\cos\left(x\right)}{\sin\left(x\right)}\right)=\] \[y\ln\left(\sin\left(x\right)\right)+\frac{xy\cos\left(x\right)}{\sin\left(x\right)}\]
4. Substitute the original equation in for \(y\) \[y'=\left(\sin\left(x\right)^{x}\right)\ln\left(\sin\left(x\right)\right)+\frac{x\left(\sin\left(x\right)^{x}\right)\cos\left(x\right)}{\sin\left(x\right)}\]
5. Evaluate the derivative at \(x=\frac{\pi}{2}\) \[\left.\frac{d}{dx}\right|_{x=\frac{\pi}{2}}\left(\sin\left(x\right)^{x}\right)\ln\left(\sin\left(x\right)\right)+\frac{x\left(\sin\left(x\right)^{x}\right)\cos\left(x\right)}{\sin\left(x\right)}=\] \[\left(\sin\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}}\right)\ln\left(\sin\left(\frac{\pi}{2}\right)\right)+\frac{\frac{\pi}{2}\left(\sin\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}}\right)\cos\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)}=\] \[\left(1^{\frac{\pi}{2}}\right)\left(\ln\left(1\right)\right)+\frac{\frac{\pi}{2}\left(1^{\frac{\pi}{2}}\right)\left(0\right)}{1}=0+0=0\]
6. Write the equation of the tangent line \[y-1=0\left(x-\frac{\pi}{2}\right)\] \[y=1\]