1. If [tex]f(x)=(x+1)^4[/tex], then [tex]f'(x)=4(x+1)^3[/tex]. So [tex]f'(1)=32[/tex].
2. With [tex]x^2+y^2=1[/tex], we differentiate once with respect to [tex]x[/tex] and get
[tex]\dfrac{\mathrm d}{\mathrm dx}[x^2+y^2]=\dfrac{\mathrm d}{\mathrm dx}1[/tex]
[tex]2x+2y\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac xy[/tex]
Differentiate again with respect to [tex]x[/tex] and we get
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{y-x\frac{\mathrm dy}{\mathrm dx}}{y^2}[/tex]
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{y+\frac{x^2}y}{y^2}=-\dfrac{y^2+x^2}{y^3}=-\dfrac1{y^3}[/tex]
(where [tex]y\neq0[/tex]).
3. Check the one-side limits where the pieces are split. For [tex]f[/tex] to be continuous everywhere, we need
[tex]\displaystyle\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=f(-1)[/tex]
[tex]\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)[/tex]
In the first case, we have
[tex]\displaystyle\lim_{x\to-1^-}f(x)=\lim_{x\to-1}x+2=1[/tex]
[tex]\displaystyle\lim_{x\to-1^+}f(x)=\lim_{x\to-1}x^2=1[/tex]
and [tex]f(-1)=1[/tex], so it's continuous here.
In the second case, we have
[tex]\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^2=1[/tex]
[tex]\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}3-x=2[/tex]
so [tex]f[/tex] is discontinuous at [tex]x=1[/tex].
4. If [tex]f(x)=3xe^x[/tex], then [tex]f'(x)=3e^x+3xe^x=3e^x(1+x)[tex]. So [tex]f'(0)=3[/tex].
5. If [tex]f(x)=(x+1)^2(x+2)^3[/tex], then [tex]f'(x)=2(x+1)(x+2)^3+3(x+1)^2(x+2)^2=(x+1)(x+2)^2(5x+7)[/tex]. So [tex]f'(0)=28[/tex].
6. The average velocity over [1, 2] is given by
[tex]\dfrac{s(2)-s(1)}{2-1}=(2^2+2)-(1^2+1)=4[/tex]
7. If [tex]f(x)=\sin^2x[/tex], then [tex]f'(x)=2\sin x\cos x=\sin2x[/tex]. So [tex]f'\left(\dfrac\pi4\right)=\sin\dfrac\pi2=1[/tex].
8. If [tex]f(x)=\log_23x[/tex], then
[tex]2^{f(x)}=3x\implies e^{\ln2^{f(x)}}=3x\implies e^{(\ln2)f(x)}=3x[/tex]
Differentiating, we get
[tex](\ln2)f'(x)e^{(\ln2)f(x)}=(\ln2)3xf'(x)=3\implies f'(x)=\dfrac1{(\ln2)x}[/tex]
So [tex]f'(1)=\dfrac1{\ln2}[/tex].
9. If [tex]f(x)=\dfrac1{x^2}[/tex], then [tex]f'(x)=-\dfrac2{x^3}[/tex]. So [tex]f'(1)=-2[/tex]
10. If [tex]f(x)=-\dfrac{6x}{e^x+1}[/tex], then [tex]f'(x)=-\dfrac{6(e^x+1)-6xe^x}{(e^x+1)^2}=-\dfrac{6e^x(1-x)+6}{(e^x+1)^2}[/tex]. So [tex]f'(0)=-\dfrac{12}4=-3[/tex].
