10. - / 2 points this is section 3.1 problem 42: for y = f(x)=xe^{x - 5}, when x = 5 and dx = 0.1: dy=…

Explanation:

Step1: Differentiate (y = xe^{x - 5}) using product - rule

The product - rule states that if (y=uv), where (u = x) and (v=e^{x - 5}), then (y^\prime=u^\prime v+uv^\prime). We know that (u^\prime = 1) and (v^\prime=e^{x - 5}). So (y^\prime=(1)\times e^{x - 5}+x\times e^{x - 5}=(x + 1)e^{x - 5}).

Step2: Calculate (dy)

We know that (dy=y^\prime dx). Substitute (x = 5) and (dx=0.1) into the formula. When (x = 5), (y^\prime=(5 + 1)e^{5 - 5}=6\times1 = 6). Then (dy=y^\prime dx=6\times0.1 = 0.6).

Step3: Calculate (f(5))

Substitute (x = 5) into (y = xe^{x - 5}), we get (f(5)=5\times e^{5 - 5}=5\times1 = 5).

Step4: Calculate the linear approximation

The linear approximation formula is (f(x+\Delta x)\approx f(x)+dy). Here (x = 5), (\Delta x=0.1), so (f(5.1)\approx f(5)+dy=5 + 0.6=5.6).

Answer:

(dy = 0.6) (f(5.1)\approx5.6)