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

$\dfrac{\mathrm d}{\mathrm dx}[x^2+y^2]=\dfrac{\mathrm d}{\mathrm dx}1$

$2x+2y\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac xy$

Differentiate again with respect to $x$ and we get

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{y-x\frac{\mathrm dy}{\mathrm dx}}{y^2}$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{y+\frac{x^2}y}{y^2}=-\dfrac{y^2+x^2}{y^3}=-\dfrac1{y^3}$

(where $y\neq0$).

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

$\dfrac{s(2)-s(1)}{2-1}=(2^2+2)-(1^2+1)=4$

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

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

$(\ln2)f'(x)e^{(\ln2)f(x)}=(\ln2)3xf'(x)=3\implies f'(x)=\dfrac1{(\ln2)x}$

So $f'(1)=\dfrac1{\ln2}$.

9. If $f(x)=\dfrac1{x^2}$, then $f'(x)=-\dfrac2{x^3}$. So $f'(1)=-2$

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