Question

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

60.\(y = x{e^{ - {x^{\bf{2}}}}}\), \(\left( {{\bf{0}},{\bf{0}}} \right)\)

Short answer

The equation of the tangent line is \(y = x\).

Step-by-step solution

Step 1:Find the value of \(y'\)

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

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x{e^{ - {x^2}}}} \right)\\ &= x{e^{ - {x^2}}}\left( {\frac{{\rm{d}}}{{{\rm{d}}x}}\left( { - {x^2}} \right)} \right) + {e^{ - {x^2}}}\left( 1 \right)\\ &= x{e^{ - {x^2}}}\left( { - 2x} \right) + {e^{ - {x^2}}}\\ &= {e^{ - {x^2}}}\left( { - 2{x^2} + 1} \right)\end{aligned}\)

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

The slope is shown below:

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

Find the equation forthe tangent

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

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

So, the equation of tangent is \(y = x\).