hw14 the chain rule (target c4; §3.6)\nscore: 6/11 answered: 6/11\nquestion 7\nfind the derivative of…

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)