Question

find the equation of the line that is tangent to the curve y = e^x sin x at the point (0, 0). equation of the tangent line: submit answer next item

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:

\(y=x\)