1.(a) Since y=2x^(3/2)+e^100, dy/dx=2*3/2*x^(1/2)+0
(b) product rule:1/x*cos(1-5x^2)+ln(x)*(-sin(1-5x^2))*(-10x)
(c) y=(4x-x^3)^(-1/2), so dy/dx=(-1/2)*(4x-x^3)^(-3/2)*(4-3x^2)
(d) Take d/dx of both sides: e^(2yx)*(2y'*x+2y)+2y*y'*x+y^2=0, solve for y': y'=(-y^2-2y*e^(2yx))/(2y*x+2*e^(2yx))
(e) chain rule: dy/dx= 1/(1+(sin(2x)+1/x)^2)*(cos(2x)*2-1/x^2)
2.(a)dy/dx=3x^2-3 at x=1 gives m=0, so y-(-2)=0*(x-1).
(b) T(x) is a line, so T(x) is a polynomial of degree 1. So T'(x) is a constant, and T''(x) is 0. So the limit is 0.
3.(a) Use Lhospitals rule to get lim_{x->0} 1/(2e^(2x))=1/2.
(b) The tangent line at x=0 is y=g(0)=g'(0)*(x-0), we have g(0)=1/2, and must find g'(0). But g'(0)=lim_{h->0}(g(h)-g(0)}/h=
lim_{h->0}(h/(e^(2h)-1)-1/2)/h
=lim_{h->0}(2h+1-e^(2h))/(2h(e^(2h-1))
=lim_{h->0}(2-2*e^(2h))/(4h*E^(2h)+2e^(2h)-2)
=lim_{h->0}(-4*e^(2h))/(8*e^(2h)+8*h*e^(2h))=-1/2
so we get 1/2-1/2*x. We can also find g'(0) by finding g'(x) and taking the limit as x->0.
4.(a) Using either the defn of the derivative or LHopital's rule we get the derivative of sin(x)^3, which is 3*sin(x)^2*cos(x).
(b) Compare rates of growth at x=\infty. ln(x^9+11) is slower than -x^6, and sin(e^x) does not grow at all. So the numerator is dominated by -x^6. For the denominator, e^(-2x) goes to 0, ln(e^(2x))=2x is small compared to 2x^6, so we get 2*x^6. The quotient then approaches -1/2.
5. (a) False. A function can decrease and still be concave upward, think of a parabola pointing up, say y=(x-1)^2.
(b) False. f(x) could have cusps at places other than x=1, for example f(x)=|x| is not diff at x=0 but is continuous everywhere.
(c) False. We know that f'(g(x))*g'(x)=1, put x=2 to find g'(2). But g(2)=1 since f(1)=2. So g'(2)=1/f'(1)=1/(1+3x^2) at x=1 which is 1/4, not 1/13.
(d) True. Note that x*p(x^2) is a polynomial with positive coeffs, so it increases starting at x=0, and approaches infinity as x-> infinity. So x*p(x^2) must attain 17 for some positive x. If x is negative, x^2 is positive, so p(x^2)>0, and x*p(x^2)<0, so no negative x could make x*p(x^2) =17. Thus there is exactly one solution to x*p(x^2)=17.
Another proof is to note that x*p(x^2) is an increasing function of x, since the derivative is a polynomial in x^2 with positive coefficients. So x*p(x^2) is an increasing function of x, so can hit 17 exactly once. Since the limiting values at infinity and -infinity are infinity and -infinity, the intermediate value theorem says 17 is hit at least once.
6. Draw any graph which decreases on (-\infty,-2), then increases on (-2,1), decreases on (1,\infty). You should have a local min at x=1, x=-2. You may or may not have several inflection points.
7. (a) (s(3)-s(0))/3=(550*3-30-25*cos(14*Pi/8))/3.
(b) s'(1), so we find s'(t)=550-25*(-sin(Pi*(5t-1)/8)*5*Pi/8 for t>1/5. Just put t=1.
(c) The left hand derivative is 2750*t at t=1/5 is 550. The right hand derivative at t=1/5 is (see (b)) 550-25*(-sin(0))*5*Pi/8=550. Since these are equal, s(t) is differentiable at t=1/5.
8. Let x be the distance in the northern direction the boat has gone, so tan(\theta)=x/1000, and dx/dt=22 ft/sec. Taking d/dt of both sides, sec^2(\theta)*d\theta/dt=1/1000*dx/dt. When the boat is 2000 feet way, sec(\theta)=2, so
d \theta/dt= 1/1000*22/4.
9. We minimize the cost C=(0.04)*Pi*r^2+(0.04)*Pi*r^2+(0.01)*2*Pi*r*h subject to 200=V=Pi*r^2*h. Substituting for h we have
C(r)=(0.08)*Pi*r^2+(0.02)*Pi*r*200/(Pi*r^2)=
=(0.08)*Pi*r^2+4/r with domain r in (0,\infty).
C'(r)=(0.16)*Pi*r-4/r^2=0 iff r=cube root of 25/Pi=about 2.
The endpoints cannot give minima, since C(r)->infty if r->0 or \infty. This is a local min since C''(r)=4/r^3, so C is concave upward on (0,\infty). So it is the absolute minimum. In this case h=200/(Pi*r^2).
10.A. f'(x)=1-a/x^2=0 iff x=\sqrt(a), which does lie on th einterval [1,a] since a>1. So the absolute min either occurs the endpoints, f(1)=a+1=f(a), or this posible local min, f(\sqrt(a))=2*\sqrt(a). Since f''(x)=2a/x^3>0 on [1,a], f(x) is concave upward on [1,a], so 2*\sqrt is the local min and absolute min too.
10B. If x is slightly less than a, 1/(a-x) is very large, and increases as x gets closer to a.
For any positive number M there is a positive number \delta such that if
a-\delta