\[\frac{d}{dx}\left[3^{2x^{4}}\right]=\frac{d}{dx}\left[2x^{4}\right]\cdot\ln\left(3\right)\cdot3^{2x^{4}}=\]

\[\left(\ln3\right)8x^{3}\left(3^{2x^{4}}\right)\]

\[\frac{d}{dx}\left[e^{\cos\left(x^{3}\right)}\right]=\frac{d}{dx}\left[\cos\left(x^{3}\right)\right]\cdot e^{\cos\left(x^{3}\right)}=\]

Use the chain rule to solve the remaining derivative

\[-\frac{d}{dx}\left(x^{3}\right)\sin\left(x^{3}\right)e^{\cos\left(x^{3}\right)}=-3x^{2}\sin\left(x^{3}\right)e^{\cos\left(x^{3}\right)}=\] \[-3x^{2}\sin\left(x^{3}\right)e^{\cos\left(x^{3}\right)}\]

\[\frac{d}{dx}\left[\ln\left(\sin\left(x^{2}\right)\right)\right]=\frac{1}{\sin\left(x^{2}\right)}\cdot\frac{d}{dx}\left(\sin\left(x^{2}\right)\right)=\] \[\frac{1}{\sin\left(x^{2}\right)}\cdot2x\cos\left(x^{2}\right)=\frac{2x\cos\left(x^{2}\right)}{\sin\left(x^{2}\right)}=\] \[\frac{2x\cos\left(x^{2}\right)}{\sin\left(x^{2}\right)}\]

Solve for the derivative by using the quotient rule, exponential and logarithmic rules, and the chain rule.

\[y'=\frac{\frac{d}{dx}\left[\ln x^{2}\right]e^{\sin\left(x\right)}-\frac{d}{dx}\left[e^{\sin\left(x\right)}\right]\ln x^{2}}{\left[e^{\sin\left(x\right)}\right]^{2}}=\] \[\frac{\frac{d}{dx}\left[x^{2}\right]\cdot\frac{1}{x^{2}}\cdot e^{\sin\left(x\right)}-\frac{d}{dx}\left[\sin\left(x\right)\right]\cdot e^{\sin\left(x\right)}\cdot\ln x^{2}}{e^{2\sin\left(x\right)}}=\] \[\frac{\frac{2x}{x^{2}}e^{\sin\left(x\right)}-\cos\left(x\right)e^{\sin\left(x\right)}\ln x^{2}}{e^{2\sin\left(x\right)}}=\frac{\frac{2}{x}-\cos\left(x\right)\ln x^{2}}{e^{\sin\left(x\right)}}\] \[\frac{\frac{2}{x}-\cos\left(x\right)\ln x^{2}}{e^{\sin\left(x\right)}}=e^{-\sin\left(x\right)}\left(\frac{2}{x}-2\cos\left(x\right)\ln x\right)\]

The slope of the given line is \(2\); therefore, we need to find the point which the derivative is equal to \(2\)

1. Find the derivative \[y'=\frac{d}{dx}\left[12x-2\right]\cdot\left(\frac{1}{6}\right)e^{12x-2}=\frac{12}{6}e^{12x-2}=\] \[2e^{12x-2}\]
2. Set the derivative equal to \(2\) \[2e^{12x-2}=2\] \[e^{12x-2}=1\] \[\ln\left(e^{12x-2}\right)=\ln\left(1\right)\] \[\left(12x-2\right)\ln\left(e\right)=0\] \[12x-2=0\] \[x=\frac{1}{6}\]

The tangent line to the graph at \(x=\frac{1}{6}\) is parallel to the line \(y=2x+5\)

Since we are asked for \(\frac{d^{n}y}{dx^{x}}\), we should look for a pattern in the first few derivatives that will continue for all subsequent derivatives.

1. Find the first derivative \[\frac{dy}{dx}=2e^{2x}\]
2. Find the second derivative without further simplifying your answer \[\frac{d^{2}y}{dx^{2}}=2\left(2e^{2x}\right)\]
3. Find the third derivative without further simplifying your answer \[\frac{d^{3}y}{dx^{3}}=2\left(2\right)\left(2e^{2x}\right)\]

As we can see from above, each subsequent derivative is multiplied by \(2\). \[\therefore\] \[\frac{d^{n}y}{dx^{n}}=2^{n}e^{2x}\]

1. Use logarithmic properties to simplify the problem \[f\left(x\right)=\log_{\pi}\left(\pi^{2^{x^{2}}}\right)=2^{x^{2}}\log_{\pi}\left(\pi\right)=2^{x^{2}}\]
2. Find the value of the function at \(x=1\) \[f\left(1\right)=2^{\left(1\right)^{2}}=2\]
3. Take the derivative \[f'\left(x\right)=\frac{d}{dx}\left(x^{2}\right)\ln\left(2\right)\left(2^{x^{2}}\right)=\ln\left(2\right)2x\left(2^{x^{2}}\right)\]
4. Evaluate the derivative at \(x=1\) \[f'\left(1\right)=\left(\ln\left(2\right)\right)\left(2\right)\left(2\right)=4\ln\left(2\right)\]
5. Write the equation of the tangent line \[y-2=4\ln\left(2\right)\left(x-1\right)\] \[\mbox{or}\] \[y=4x\ln\left(2\right)-4\ln\left(x\right)+2\]