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

find an equation of the line tangent to the following equation at point ( x = 1 ).\n( y = x^{cos x} )\nthe equation of the tangent line is (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( y = x^{cos x} )\nthe equation of the tangent line is (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^{\cos x}), take (\ln) on both sides: (\ln y=\cos x\ln x).

Step2: Differentiate both sides with respect to (x)

Using the chain - rule ((\ln y)^\prime=\frac{y^\prime}{y}) and the product - rule ((uv)^\prime = u^\prime v+uv^\prime) (where (u = \cos x), (v=\ln x), (u^\prime=-\sin x), (v^\prime=\frac{1}{x})). We get (\frac{y^\prime}{y}=-\sin x\ln x+\frac{\cos x}{x}). Then (y^\prime=y\left(-\sin x\ln x+\frac{\cos x}{x}\right)). Since (y = x^{\cos x}), (y^\prime=x^{\cos x}\left(-\sin x\ln x+\frac{\cos x}{x}\right)).

Step3: Find the value of (y) and (y^\prime) at (x = 1)

When (x = 1):

  • For (y): (y(1)=1^{\cos1}=1).
  • For (y^\prime): (y^\prime(1)=1^{\cos1}\left(-\sin1\ln1+\frac{\cos1}{1}\right)=\cos1).

Step4: Use the point - slope form of a line (y - y_0=m(x - x_0))

Here (x_0 = 1), (y_0 = 1), (m=\cos1). The equation is (y - 1=\cos1(x - 1)). Expand it: (y=\cos1\cdot x+(1 - \cos1)).

Answer:

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