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)})