Question

57-60 Find an equation of a tangent line to the curve at the given point.

57. \(y = {{\bf{2}}^x}\), \(\left( {{\bf{0}},{\bf{1}}} \right)\)

Short answer

The equation of a tangent line is \(y = \left( {\ln 2} \right)x + 1\).

Step-by-step solution

Step 1:Find the value of \(y'\)

Find the value of \(y'\) for the equation\(y = {2^x}\).

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{2^x}} \right)\\ &= {2^x}\ln 2\end{aligned}\)

Find the slope of the tangent at point \(\left( {{\bf{0}},{\bf{1}}} \right)\)

The slope is shown below:

\(\begin{aligned}{\left. {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right|_{\left( {0,1} \right)}} &= {2^0}\ln 2\\ &= \ln 2\end{aligned}\)

Find the equation forthe tangent

The equation of tangent passing through \(\left( {0,1} \right)\) with slope \(\ln 2\) is:

\(\begin{aligned}y - 1 &= \ln 2\left( {x - 0} \right)\\y &= \left( {\ln 2} \right)x + 1\end{aligned}\)

So, the equation of tangent is \(y = \left( {\ln 2} \right)x + 1\).