Question

hw14 the chain rule (target c4; §3.6)
score: 6/11 answered: 6/11
question 7
find the derivative of: 2e^{-3x}cos(7x). hint: use product rule and chain rule
use e^x for e^x.
now, find the equation of the tangent line to the curve at x = 0. write your answer in mx + b format.
y=
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Explanation:

Step1: Recall product - rule

The product - rule states that if \(y = uv\), where \(u\) and \(v\) are functions of \(x\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = 2e^{-3x}\) and \(v=\cos(7x)\).

Step2: Find \(u^\prime\) using chain - rule

If \(u = 2e^{-3x}\), let \(t=-3x\), then \(\frac{du}{dt}=2e^{t}\) and \(\frac{dt}{dx}=-3\). By the chain - rule \(\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}\), so \(u^\prime=2e^{-3x}\cdot(-3)=-6e^{-3x}\).

Step3: Find \(v^\prime\) using chain - rule

If \(v = \cos(7x)\), let \(s = 7x\), then \(\frac{dv}{ds}=-\sin(s)\) and \(\frac{ds}{dx}=7\). By the chain - rule \(\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}\), so \(v^\prime=-7\sin(7x)\).

Step4: Apply product - rule

\(y^\prime=u^\prime v+uv^\prime=-6e^{-3x}\cos(7x)-14e^{-3x}\sin(7x)=-2e^{-3x}(3\cos(7x) + 7\sin(7x))\).

Step5: Find the slope of the tangent line at \(x = 0\)

Substitute \(x = 0\) into \(y^\prime\): \(y^\prime(0)=-2(3\cos(0)+7\sin(0))=-6\).

Step6: Find the value of \(y\) at \(x = 0\)

Substitute \(x = 0\) into \(y = 2e^{-3x}\cos(7x)\): \(y(0)=2e^{0}\cos(0)=2\).

Step7: Find the equation of the tangent line

The equation of a line is \(y - y_0=m(x - x_0)\), where \((x_0,y_0)=(0,2)\) and \(m=-6\). So \(y-2=-6(x - 0)\), which simplifies to \(y=-6x + 2\).

Answer:

The derivative is \(-2e^{-3x}(3\cos(7x)+7\sin(7x))\) and the equation of the tangent line is \(y=-6x + 2\)