Question

find the equation of the normal to the graph of ( y = x^{x} ) at the point ( (2,4) ).

Explanation:

Step1: Find the derivative of \(y = x^{x}\)

Let \(y=x^{x}\), take the natural logarithm on both sides: \(\ln y=x\ln x\).
Differentiate both sides with respect to \(x\):
\(\frac{1}{y}y'=\ln x + 1\), so \(y'=x^{x}(\ln x + 1)\).

Step2: Evaluate the derivative at \(x = 2\)

When \(x = 2\), \(y'=2^{2}(\ln 2+ 1)=4(1+\ln 2)\).

Step3: Find the slope of the normal line

The slope of the tangent line at \(x = 2\) is \(m_{t}=4(1 +\ln 2)\).
The slope of the normal line \(m_{n}\) satisfies \(m_{t}\times m_{n}=-1\), so \(m_{n}=-\frac{1}{4(1+\ln 2)}\).

Step4: Use the point - slope form to find the equation of the normal line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(2,4)\) and \(m =-\frac{1}{4(1+\ln 2)}\).
\(y - 4=-\frac{1}{4(1+\ln 2)}(x - 2)\)
\(y=-\frac{1}{4(1+\ln 2)}x+\frac{1}{2(1+\ln 2)}+4\)
\(y=-\frac{1}{4(1+\ln 2)}x+\frac{1 + 8(1+\ln 2)}{2(1+\ln 2)}\)
\(y=-\frac{1}{4(1+\ln 2)}x+\frac{9 + 8\ln 2}{2(1+\ln 2)}\)

Answer:

\(y =-\frac{1}{4(1+\ln 2)}x+\frac{9 + 8\ln 2}{2(1+\ln 2)}\)