By implicit differentiation:
<span>(x(dy/dx) + y)e^(xy) = 0 </span>
<span>Note that when differentiating e^(xy), apply chain rule. When differentiating xy, use product rule. Also: When differentiating y w/respect to x, think of that as if you are differentiating f(x). </span>
<span>Then, substitute (1,ln(2)) and solve for dy/dx. </span>
<span>(1(dy/dx) + ln(2))e^(1ln(2)) = 0 </span>
<span>((dy/dx) + ln(2))e^(ln(2)) = 0 </span>
<span>Note that e^(ln(2)) = 2 since e and ln are inverse of each other. </span>
<span>2((dy/dx) + ln(2)) = 0 </span>
<span>dy/dx + ln(2) = 0 . . . . You get this expression by dividing both sides by 2 </span>
<span>dy/dx = -ln(2) . . . . . . .Subtract both sides by ln(2) </span>
<span>Therefore, dy/dx = -ln(2) </span>
<span>I hope this helps!</span>
Answer:
1 serving is 30 crackers
2 servings is 60 crackers
3 servings is 90 crackers
4 servings is 120 crackers
5 servings is 150 crackers.
Step-by-step explanation:
To find out how many crackers for how many servings you multiply one serving (30 crackers) by the amount of servings.
Example:
1 serving 30 × 1 = 30
2 servings 30 × 2 = 60
3 servings 30 × 3 = 90
F(x)=3x/2 for 0≤x≤2
<span>.....=6 - 3x/2 for 2<x≤4 </span>
<span>g(x) = -x/4 + 1 for 0≤x≤4 and g'(x)=-1/4 </span>
<span>so h(x)= f(g(x)) = (3/2)(-¼x+1)=-3x/8 + 3/2 for 0≤x≤2 </span>
<span>for x=1, h'(x)=-3/8 so h'(1)=-3/8 </span>
<span>When x=2, g(2)=1/2 so h'(2)=g'(2)f '(1/2)= -(1/4)(3/2)=-3/8 </span>
<span>When x=3, h(x)=6 - (3/2)(1 - x/4) = 9/2 +3x/8 </span>
<span>h'(x)=3/8 so h'(3) = 3/8</span>
