Question

Can anyone help me with calculus??

Answer 1

1. If
`f(x)=(x+1)^4`, then
`f'(x)=4(x+1)^3`. So
`f'(1)=32`.

2. With
`x^2+y^2=1`, we differentiate once with respect to
`x` and get

`(\mathrm d)/(\mathrm dx)[x^2+y^2]=(\mathrm d)/(\mathrm dx)1`

`2x+2y(\mathrm dy)/(\mathrm dx)=0`

`(\mathrm dy)/(\mathrm dx)=-\frac xy`

Differentiate again with respect to
`x` and we get

`(\mathrm d^2y)/(\mathrm dx^2)=-(y-x(\mathrm dy)/(\mathrm dx))/(y^2)`

`(\mathrm d^2y)/(\mathrm dx^2)=-\frac{y+\frac{x^2}y}{y^2}=-(y^2+x^2)/(y^3)=-\frac1{y^3}`

(where
`y\\eq0`).

3. Check the one-side limits where the pieces are split. For
`f` to be continuous everywhere, we need

`\displaystyle\lim_(x\to-1^-)f(x)=\lim_(x\to-1^+)f(x)=f(-1)`

`\displaystyle\lim_(x\to1^-)f(x)=\lim_(x\to1^+)f(x)=f(1)`

In the first case, we have

`\displaystyle\lim_(x\to-1^-)f(x)=\lim_(x\to-1)x+2=1`

`\displaystyle\lim_(x\to-1^+)f(x)=\lim_(x\to-1)x^2=1`

and
`f(-1)=1`, so it's continuous here.

In the second case, we have

`\displaystyle\lim_(x\to1^-)f(x)=\lim_(x\to1)x^2=1`

`\displaystyle\lim_(x\to1^+)f(x)=\lim_(x\to1)3-x=2`

so
`f` is discontinuous at
`x=1`.

4. If
`f(x)=3xe^x`, then
`f'(x)=3e^x+3xe^x=3e^x(1+x)[tex]. So [tex]f'(0)=3`.

5. If
`f(x)=(x+1)^2(x+2)^3`, then
`f'(x)=2(x+1)(x+2)^3+3(x+1)^2(x+2)^2=(x+1)(x+2)^2(5x+7)`. So
`f'(0)=28`.

6. The average velocity over [1, 2] is given by

`(s(2)-s(1))/(2-1)=(2^2+2)-(1^2+1)=4`

7. If
`f(x)=\sin^2x`, then
`f'(x)=2\sin x\cos x=\sin2x`. So
`f'\left(\frac\pi4\right)=\sin\frac\pi2=1`.

8. If
`f(x)=\log_23x`, then

`2^(f(x))=3x\implies e^{\ln2^(f(x))}=3x\implies e^((\ln2)f(x))=3x`

Differentiating, we get

`(\ln2)f'(x)e^((\ln2)f(x))=(\ln2)3xf'(x)=3\implies f'(x)=\frac1{(\ln2)x}`

So
`f'(1)=\frac1{\ln2}`.

9. If
`f(x)=\frac1{x^2}`, then
`f'(x)=-\frac2{x^3}`. So
`f'(1)=-2`

10. If
`f(x)=-(6x)/(e^x+1)`, then
`f'(x)=-(6(e^x+1)-6xe^x)/((e^x+1)^2)=-(6e^x(1-x)+6)/((e^x+1)^2)`. So
`f'(0)=-\frac{12}4=-3`.