find an equation of the line tangent to the following equation at point ( x = 1 ).\n\n( y = x ^ { cot x }…

find an equation of the line tangent to the following equation at point ( x = 1 ).\n\n( y = x ^ { cot x } )\n\nthe equation of the tangent line is \n(type an equation using ( x ) and ( y ) as the variables. type an exact answer.)

find an equation of the line tangent to the following equation at point ( x = 1 ).\n\n( y = x ^ { cot x } )\n\nthe equation of the tangent line is \n(type an equation using ( x ) and ( y ) as the variables. type an exact answer.)

Answer

Explanation:

Step1: Take the natural logarithm of both sides

Given (y = x^{\cot x}), take (\ln) on both sides: (\ln y=\cot x\ln x).

Step2: Differentiate both sides with respect to (x)

Using the chain - rule on the left side (\frac{d}{dx}(\ln y)=\frac{1}{y}y') and the product - rule on the right side (\frac{d}{dx}(\cot x\ln x)=-\csc^{2}x\ln x+\frac{\cot x}{x}). So, (y'=y\left(-\csc^{2}x\ln x+\frac{\cot x}{x}\right)).

Step3: Find the value of (y) at (x = 1)

Substitute (x = 1) into (y=x^{\cot x}). Since (\cot(1)=\frac{\cos(1)}{\sin(1)}) and (y(1)=1^{\cot(1)} = 1).

Step4: Find the slope (m) of the tangent line at (x = 1)

Substitute (x = 1) into (y'). (\csc(1)=\frac{1}{\sin(1)}), (\cot(1)=\frac{\cos(1)}{\sin(1)}), (y'(1)=1\times\left(-\csc^{2}(1)\ln(1)+\frac{\cot(1)}{1}\right)). Since (\ln(1) = 0), (y'(1)=\cot(1)=\frac{\cos(1)}{\sin(1)}).

Step5: Use the point - slope form (y - y_1=m(x - x_1))

Here (x_1 = 1), (y_1 = 1), (m=\cot(1)). The equation is (y-1=\cot(1)(x - 1)), which simplifies to (y=\cot(1)x+(1-\cot(1))).

Answer:

(y=\cot(1)x+(1 - \cot(1)))