\(\displaystyle \begin{align*}&\text{Our function, }17cos(16sin(6x))\text{, is one function nested within another.}\\&\text{Note the }16sin(6x)\text{ inside of the }17cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&u=16sin(6x)\\&du=96cos(6x)\\&f'(x)=(96cos(6x))(-17sin(16sin(6x)))=-1632sin(16sin(6x))cos(6x)\end{align*}\)

\(\displaystyle \begin{align*}&\text{Our function, }-17sin(15sin(x))\text{, is one function nested within another.}\\&\text{Note the }15sin(x)\text{ inside of the }-17sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&u=15sin(x)\\&du=15cos(x)\\&f'(x)=(15cos(x))(-17cos(15sin(x)))=-255cos(15sin(x))cos(x)\end{align*}\)

\(\displaystyle \begin{align*}&\text{The function }f(x)=20e^{(5tan(4x))}\text{, can be seen as a function nested within another.}\\&\text{Notice the }5tan(4x)\text{ within the }20e^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&u=5tan(4x)\\&du=20tan(4x)^{2} + 20\\&f'(x)=(20tan(4x)^{2} + 20)(20e^{(5tan(4x))})=20e^{(5tan(4x))}\cdot (20tan(4x)^{2} + 20)\end{align*}\)

\(\displaystyle \begin{align*}&\text{The function }f(x)=19tan(17cos(5x))\text{, can be seen as a function nested within another.}\\&\text{Notice the }17cos(5x)\text{ within the }19tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=17cos(5x)\\&du=-85sin(5x)\\&f'(x)=(-85sin(5x))(19tan(17cos(5x))^{2} + 19)=-1615sin(5x)\cdot (tan(17cos(5x))^{2} + 1)\end{align*}\)

\(\displaystyle \begin{align*}&\text{Our function, }-6sin(4cos(3x))\text{, is one function nested within another.}\\&\text{Note the }-4cos(3x)\text{ inside of the }6sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&u=-4cos(3x)\\&du=12sin(3x)\\&f'(x)=(12sin(3x))(6cos(4cos(3x)))=72sin(3x)cos(4cos(3x))\end{align*}\)

\(\displaystyle \begin{align*}&\text{The function }y=-cos(10tan(5x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }10tan(5x)\text{ is inside of the }-1cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&u=10tan(5x)\\&du=50tan(5x)^{2} + 50\\&\frac{dy}{dx}=(50tan(5x)^{2} + 50)(sin(10tan(5x)))=sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)\end{align*}\)

\(\displaystyle \begin{align*}&\text{The function }y=4cos(7sin(x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-7sin(x)\text{ is inside of the }4cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&u=-7sin(x)\\&du=-7cos(x)\\&\frac{dy}{dx}=(-7cos(x))(4sin(7sin(x)))=-28sin(7sin(x))cos(x)\end{align*}\)

\(\displaystyle \begin{align*}&\text{Our function, }10cos(2\cdot 3^x)\text{, is one function nested within another.}\\&\text{Note the }-2\cdot 3^x\text{ inside of the }10cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&u=-2\cdot 3^x\\&du=-2\cdot 3^xln(3)\\&f'(x)=(-2\cdot 3^xln(3))(10sin(2\cdot 3^x))=-20\cdot 3^xln(3)sin(2\cdot 3^x)\end{align*}\)

\(\displaystyle \begin{align*}&\text{The function }y=5sin(3e^{(6x)})\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }3e^{(6x)}\text{ is inside of the }5sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&u=3e^{(6x)}\\&du=18e^{(6x)}\\&\frac{dy}{dx}=(18e^{(6x)})(5cos(3e^{(6x)}))=90e^{(6x)}cos(3e^{(6x)})\end{align*}\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} s}F(s)= \frac{d}{ds}\int_{0}^{\infty}e^{-st}*f(t) dt=-t*[\int_{0}^{\infty}e^{-st}*f(t) dt]=-tF(s)\)