find the equation of the line that is tangent to the curve y = e^x sin x at the point (0, 0). equation of…

Explanation:

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

Use the product - rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = e^{x}) and (v=\sin x). The derivative of (e^{x}) is (e^{x}), and the derivative of (\sin x) is (\cos x). So (y^\prime=e^{x}\sin x+e^{x}\cos x).

Step2: Evaluate the derivative at (x = 0)

Substitute (x = 0) into (y^\prime). (y^\prime(0)=e^{0}\sin(0)+e^{0}\cos(0)=0 + 1\times1=1). The value of the derivative at (x = 0) is the slope (m) of the tangent line.

Step3: Use the point - slope form of a line

The point - slope form is (y - y_{1}=m(x - x_{1})), with ((x_{1},y_{1})=(0,0)) and (m = 1). So (y-0 = 1\times(x - 0)), which simplifies to (y=x).

Answer:

(y=x)