Calculate \(y'\) for each of the following functions.
  • If \(y=\sqrt {2x+1}\), then \(y' = \answer {\frac {1}{\sqrt {2x+1}}}\).
  • If \(y=e^x+x^e\), then \(y' = \answer {e^x+ex^{e-1}}\).
  • If \(y=x\sin (\pi x)\), then \(y' = \answer {\sin (\pi x) +\pi x \cos (\pi x) }\).
  • If \(y=\ln (x^4+5)\), then \(y' = \answer {\frac {4x^3}{x^4+5}}\).
  • If \(y=\left (\frac {\cos (2x)}{x}\right )^3\), then \(y' = \answer {3\left (\frac {\cos (2x)}{x}\right )^2 \cdot \frac {-2x\sin (2x)-\cos (2x)}{x^2}}\).
Find the equation of the tangent line to the graph of \(y=3e^{2x}+1\) at the point \(x=0\). Express your final answer in the form \(y=mx+b\).

The tangent line is \(y=\answer {6x+4}\).

The slope of the tangent line is given by \(y'(0)\). Since \(y=3e^{2x}+1\),
\[ y' = \answer {6e^{2x}} \]

and thus \(y'(0) = \answer {6}\).

A point on the line is found using the function. Since \(y(0) = \answer {4}\), a point on the line is \((x,y)= \left (\answer {0},\answer {4}\right )\).

Now, use the point-slope form for a line

\[ y-y_0=m(x-x_0) \]
and solve for \(y\).